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Problems-Bodies in vertical  motion.31.A stone is projected vertically upwards with a velocity 29.4m/sec. i)Calculate the maximumheight to which it rises ii) calculate the time taen to reach maximum height iii) !ind theratio o" velocities a"ter 1sec# 2 sec# $sec o" its journey.%oln& !rom the data given in the pro'lem#

(nitial velocity o" stone u29.4 m/sec#

Acceleration due to gravity g9.* #

i)

14.+ m.

ii) ,ime o" ascent $sec.

iii) -atio o" velocities o" a vertically projected up 'ody a"ter 1 sec# 2 sec# $sec # . . . . . . . . . . .. . . o" its journey will 'e & & . . . . . . . . . . . . & ug) & u2g) & u$g) & . . . . . . . . & ung).

& & 29.49.* ) & 29.4 19.0) & 29.4 29.4)

& & 19.0 & 9.* & .

2.A stone is dropped "rom the top o" a tower.,he stone touches the ground a"ter3sec.Calculate the i)height o" the tower and the ii) velocity with which it stries the ground.%oln& !rom the data given in the pro'lem#

(nitial velocity o" the stone u m/sec#Acceleration due to gravity g 9.*latex m/sec52 ,ime o" descent 3 sec.i) 6eight o" the tower 'e %6 say)

su'stitute the a'ove values in the e7uation %ut8

6 3)8 8 4.9)23) 122.3 m.ii) et the velocity when it touches v say)#su'stitute the a'ove values in the e7uation vu 8gv 89.*)3) 49 m/sec.$.A stone is projected vertically upwards with an initial velocity 9* m/sec . Calculate i)maximum height the 'ody reaches ii ) :nd its velocity when it is exactly at the mid point o"it;s journey iii) time o" <ight.%oln & !rom the data given in the pro'lem#(nitial velocity o" the stone u 9* m/sec#

acceleration due to gravity a 9.* #

i) =aximum height 49 m.

ii) >hen it is in the middle % 243 met the velocity v say)#

su'stitute the values in the e7uation 2g%#

we get 29.*)243)#

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  904 4*2 4*2 #v 09.$ m/sec

iii) time o" <ight , 2 sec.4.?stimate the "ollowing.a) how long it too @ing @ong to "all straight down "rom the top o" a $0 m high 'uilding

 BBBBBBBBB seconds') his velocity just 'e"ore landingDBBBBBBBBBBB m/s.  Euestion 'y ?(%? in Fahoo answers).%oln& !rom the data given in the pro'lem#

6eight o" 'uilding %6 $0m#

Acceleration Gue to gravity g1 #

i) time o" descent 'e

su'stitute the values in the "ormula %ut8

we get $0 ) ) 8 #

$0 3 H $0/3+2

 ,here"ore *.49 sec.

ii )Ielocity o" ing ong just 'e"ore touching the ground v g #

v 1)*.49) *4.9 m/sec.

3. A 'ase'all is hit straight up into the air with a speed o" 2$ m/s.a) 6ow high does it goBBBB m') 6ow long is it in the airBBBBB s.  Euestion 'y ?(%? in Fahoo answers).%oln& !rom the data given in the pro'lem#

Acceleration Gue to gravity g1 #

a) =aximum height 

329/2 20.43 m.

' ) ,ime o" <ight o" 'ase 'all , #

 , # 40/1 4.0 sec.

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0.A 'ody starting "rom rest slides down an inclined plane.!ind the velocity a"ter it has

descended vertically a distance o" 3metres. g9.* ).%oln& ,he velocity o" the 'ody when it touches the ground sliding down an inclined plane #will'e same as when the 'ody vertically "alls "reely "rom height JhD.

!rom the pro'lem height %h3mH acceleration due to gravity g9.* H (nitial velocityu H I

su'stitute the values in the e7uation 2gs #

we get 29.*) 3) H 9*

I 9.*99 m/sec.

+.A 'all projected vertically upwards returns to ground a"ter 13sec.Calculate i)maximum

height to which it rises ii)velocity with which it is projected and iii)its position a"ter 0seconds.%oln&!rom the data given in the pro'lem#

Ielocity o" projection u

acceleration due to gravity g9.* #

time o" <ight , 13 sec

su'stitute the values o" ,#g in the e7uation ,

13 H 2u 13 9.*) 14+

ii )Ielocity with which it is projected u +$.3 m/sec.

i)=aximum height #

su'stitute the values o" u#g in the e7uation we get

2+3.023 m

1iii) et (ts position a"ter t0sec 'e %

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su'stitute the values o" u#t and g in the e7uation %ut

% +$.3)0) 4411+0.4 204.0 m high "rom the ground.

*.A stone was dropped "rom a rising 'aloon at a height o" 13m a'ove the ground and itreaches the ground in 13 seconds.!ind the velocity o" the 'aloon at the instant the stonewas dropped.%oln& !rom the data given in the pro'lem#

Acceleration due to gravity g9.* #

6eight o" the 'alloon when the stone is dropped "rom it h13m#

 ,ime o" <ight ,13 sec#

et the velocity o" the 'alloon when the stone is dropped "rom it is u say).

%u'stitute the values o" g#t and h in the e7uation h ut

we get 13 u13) 4.9)223) 13u#

13 13K4.9)13) uL#

1+$.3 u H u +$.3 1 0$.3 m/sec.

9. !rom the top o" a tower o" 2 metres high# a stone is projected vertically upwards with avelocity 49m/sec.Calculate the i)maximum height traveled 'y it "rom ground level# ii)thevelocity with which it stries the ground and the iii) time it taes to reach the ground.

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%oln& !rom the data given in the pro'lem#6eight o" the tower is

h2m#

velocity o" projection u 49 m/sec#

=ax height o" the stone "rom top o" the tower 'e % M say)#

velocity at maximum height v #

i) su'stitute the values o" u#v and g in e7uation 2gs#

we get 29.*)M#

M 122.3 m "rom the top o" the tower

=aximum height "rom ground level 6 M 8h 122.3 8 2 $22.3 m.

ii) et the velocity with which it reaches the ground 'e say#

su'stitute the values in the e7uation

#

#

+9.3 m/sec.

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iii) ,ime o" <ight ,

%u'stitute the values u#g and h in the e7uation h ut#

we get 2 49,#

2 49,

1, #

'y solving this 7uadratic e7uation "or , we get , 1$.11 sec

Problems Linear motion.(n this post and in "ew o" my posts to come# ( would lie to solve pro'lems on linearmotion#"reely "alling 'odies#vertically projected up 'odies and projectiles .

1.An o'ject accelerates "rom rest to a velocity 2m/sec in 4seconds.(" the o'ject hasuni"orm acceleration# :nd its acceleration and displacement in this time.%oln& !rom the data given in the pro'lem we have#

(nitial velocity u #

:nal velocity v2 m/sec#

 ,ime o" journey t4sec#

Acceleration a 3

Gisplacement % ut 8 4)8

% 4m.

2.An o'ject starting "rom rest moves with uni"orm acceleration o" $ "or 0sec.!ind itsvelocity and displacement a"ter 0seconds.%oln& !rom the data given in the pro'lem we have#

(nitial velocity o" the o'ject u #

Acceleration o" the o'ject a $ #

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 ,ime o" journey t 0sec.

!inal velocity o" the o'ject vu8at 8 $0) 1* m/sec.

Gisplacement % ut 8 0)8

% 34m.

$.An o'ject starting "rom rest moves with uni"orm acceleration o" 4 .!ind itsdisplacement i) 3seconds ii) in 3th second iii) *th second.%oln & !rom data given in the pro'lem

(nitial velocity o" the o'ject u#

Acceleration o" the o'ject a4 #

i) time t3 seconds#

Gisplacement o" the o'ject %ut 8 3)8

Gisplacement o" the o'ject in 3seconds % 3m.

ii) Gisplacement o" the 'ody in 3th second

et us su'stitute n3 in the "ormula u8an1/2)

8431/2) 44.3) 1*m.

iii)Gisplacement o" the 'ody in *th second

et us su'stitute n* in the "ormula u8an1/2)

84*1/2) 4+.3) $m.

4.An o'ject started moving with an initial velocity o" 1m/sec# a"ter traveling a distance o"3m gets a velocity 2m/sec.!ind its i) acceleration ii) time taen "or 3m displacement.soln& !rom the data given in the pro'lem#

(nitial velocity o" the o'ject u1m/sec#

!inal velocity o" the o'ject v 2m/sec#

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Gisplacement %3m#

i) acceleration a

%u'stitute the values o" u#v and % in the e7uation 2as#

we get 2a3)

$ 1a or a $/1$ .

ii),ime t

%u'stitute the values o" u#v and a in the e7uation vu8at#

we get 218$)t H 1$t

t 1/$ 1/$ .$$$ sec.

3.>hen an o'server started o'serving a car it;s velocity was x m/sec # i" it travels "or 1sec

with uni"orm acceleration 2.3 and its velocity increases to +3m/sec.!ind i) (nitialvelocity o" the car ii) Gisplacement o" the car in 1sec iii) Gisplacement o" the car in :rst3sec and last 3sec# what is your in"erence.%oln& !rom the data given in the pro'lem#

(nitial velocity o" the car u x say)#

!inal velocity +3m/sec#

Acceleration a 2.3 #

 ,ime o" journey t1sec.

i) %u'stitute the values o" u#v#a and t in the e7uation vu8at

we get +3 x82.31) H x3 m/sec.

ii) et the displacement o" the car in 1sec 'e %

%u'stitute the values o" u#a and t in the e7uation %ut 8

>e get % 31)8 H

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% 38123 H s0+3 m

ii) et the displacement in :rst 3sec 'e

%u'stitute t3 sec and thve values o" u and a in the e7uation ut 8

we get 33) 8

  23 8 $1.23 2*1.23 m.

et the displacement in next 3sec 'e

%

  0+3 2*1.23 $9$.+3 m

>e can o'serve that# even though the time o" journey is same # N

Gisplacement o" the 'ody in second hal" is greater than in the :rst hal" time o" its journey.

0 .A cheetah can accelerate "rom rest to 24. m/s in 0.+ s.Assuming constant acceleration# how "ar has the cheetah run in this timeEuestion 'y suOiin yahoo answers).

%oln& !rom the data given in the pro'lem#

(nitial velocity o" the cheetah u#

!inal velocity o" cheetah v24 m/sec#

 ,ime t 0.+sec

Gistance traveled 'y cheetah 'e %

 ,he e7uations o" motion are 2a% 1)

and a 2)

!rom e7uations 1) and 2) we get 2 %

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simpli"ying it we get %

%u'stitute the values o" u#v and t we get % 120.+)*.4 m

+) A car covers :rst halt o" the distance 'etween two places at a speed o" $ m/hr and thesecond hal" at a 9m/hr.>hat will 'e the average speed o" the car%oln& =ethod (& et the total distance 'etween the places 'e %.

 ,ime taen to cover !irst hal" hours.

 ,ime taen to cover %econd hal" hours.

 ,otal time t 8 8 .

Average speed ,otal distance /total time.

%/t

43m/hr.

=ethod (( %hort cut)& (" the 'ody covers 1st hal" o" distance with a speed x and the second

hal" with a speed y#then the average speed  .

Average speed latex P"racQ2+RQ0R 43m/hr.

*) A 'ody starts "rom rest and ac7ires a velocity o" 4m/sec in 1seconds.Calculate theacceleration and distance traveled.%oln& !rom the data given in the pro'lem

(nitial velocity o" the 'ody u#

 ,ime o" journey t1sec#

!inal velocity v4m/sec#

Acceleration a and distance traveled in 1sec s

%u'stitute the values o" u#v and t in the e7uation vu8at#

we get 4 )1) 8 a 1) H 1a4

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a 4

%u'stitute the values o" u#a and t in the e7uation % ut 8

%)1)8 82 2m.

9) A 'ody moving with uni"orm acceleration covers 0m in second and 10m in

second.Calculate the initial velocity#acceleration and distance moved in second .%oln& !rom the data given in the pro'lem

Gistance moved in second 0m#

Gistance moved in second 10m#

  a42) 2a

there"ore 2a 100 1

a 3 m/

%u'stitute the values o" #a and n2 in the e7uation u8an 1/2)

we get 0u8321/2) H 0u8+.3

u+.3 80 1.3 m/sec.

DISTANCE MOVED IN TE SECOND ! "#A$%-&'()*

%u'stitute the values o" u#a in the a'ove e7uation we get second1.3833.3)

 second 1.382+.3 20m.

1) An o'ject started traveling with a velocity 2m/sec moves with an uni"orm acceleration o" 

$ m/ .

i) !ind the ratio o" displacements in a) # # seconds ') # # and seconds .

ii)!ind the ratio o" velocities a) # # seconds ') # # seconds .

%oln& !rom the data given in the pro'lem

(nitial velocity u2 m/sec#

Acceleration a $ m/ #

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i a) ,he ratio o" displacements in # # seconds

!rom the "ormula & & 2u8a) & 2u83a) & 2u89a)

& & 48$) & 4813) & 482+)

& & +&19&$1 .

i') ,he ratio o" displacements in # # seconds

!rom the "ormula & & 2u8$a) & 2u8+a) & 2u811a)

& & 489) & 4821) & 48$$)

& & 1$ & 23 & $+ .

ii a) !rom the "ormula & & & . . . . . . . . & u8a) & u82a) & u8$a) & . . . . . . . .. & u8na).

 ,he ratio o" velocities a) # # seconds

& & u8a) & u8$a) & u83a) 3 &11 & 1+ .

ii ') ,he ratio o" velocities a) # # seconds

& & u82a) & u84a) & u80a) * &1 4 &2 .

11) A 'ody travels 2cm in the :rst two seconds and 22cm in the next"our seconds.>hat will 'e the velocity at the end o" the seventh second "rom the start

%oln& ,he displacement o" the 'ody in :rst 2 sec 2cm#

et the initial velocity usay) #

Acceleration asay)# time 2sec

  8 #

%u'stitute the value o" in a'ove e7uation# we get

  2u8 #

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  2u 82a H 2u8a) 2

 ,here"ore u8a 1 1)

Siven that the 'ody travels 22cm in next 4sec.,hat is "rom the start it displaces 2822

42 cm in 0sec.

Gisplacement 42 cm#

time 0 sec#

su'stitute theses values in the e7uation u 8 #

  0u8 0u81*a#

0u8$a) 42 H u8$a + 2)

%olving e7uations 1) and 2) or ?7 2) ?71)

we get 2a $ H a13 cm/ #

%u'stitute value o" a in ?71) we get u13 1#

u 113 cm/sec.

 ,he velocity at the end o" second vu8a

we get v 113813)0) 1139 23cm/sec.

 ,here"ore :nal velocity v23cm/sec.

12.A su'way train starts "rom rest at a station and accelerates at a rate o" 10.3 "or

1$.1 sec.(t runs at constant speed "or 09.+s and slows down at a rate o" $.43 until itstops at the next station.>hat is the total distance covered%oln& !rom the data given in the pro'lem#

(nitial velocity o" the train u m/sec#

Acceleration a 10.3 #

time t 1$.1 sec#

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Ielocity a"ter 1$.1 sec v

and the distance traveled

%u'stitute the values in the e7uation v u8at

we get v 810.3)1$.1) 210.13 m/sec.

%u'stitute in e7uation ut 8

we get )1$.1)8

  81413.+* 1413.+* m

A"ter that the train travels with velocity v210.13 m/sec "or 09.+sec. calculate distance

traveled during this time.

v210.13 m/sec# t 09.+ sec

su'stitute the values in e7uation vt 210.13)09.+)1303.00 m

!inally the trains decelerates at the rate o" $.43 and comes to rest.:nd distancetraveled 'e"ore coming to rest.

Acceleration a $.43 #

(nitial velocity u 210.13m/sec#

!inal velocity v#

distance traveled

su'stitute the values in 2as#

we get 2$.43)

0.9 40+2.*2

  0++1.1$.

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 ,otal distance traveled 'y train % 8 8 1413.+*81303.0080++1.1$2$232.3+m or 2$.232 m.

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 ,agged acceleration#  Answers to your Questions.# displacement #Problems

 Jan 10

+inematics1&.T,e terms est Motion are relative terms.

 ,here is no o'ject in the universe which is at absol/te rest.?x& A text 'oo in a'oo rac may 'e relatively at rest# with respect to the immediate surroundings.Uut# the earth is revolving around sun# there"ore every particle on earth will 'e inmotion including the text 'oo.6ence# the text 'oo also will 'e in motion withrespect to %un.

 ,here is no o'ject in the universe which is in absol/te motion. ?x&A persontraveling in a train will 'e in motion as the train is moving.Uut# with respect to the"ellow passengers and all the non moving o'jects in the train the person will 'erelatively at rest.

(. T,e s,ortest distance bet0een initial and 1nal 2ositions o3 a movin4 bod5 iscalled its dis2lacement.

?x& (" a o'ject starts its journey "rom a point A# travels in a circular path and againreaches the point A. ,hen its displacement is Vero#'ecause its initial and :nal pointsare same. Uut# the distance traveled is d 2 r length o" the circum"erence o"circular path# i.e the distance d) traveled 'y a o'ject may not 'e e7ual to itsdisplacement %). >hen the o'ject moves in a curved path or Oig Oag path dN%.

(" the o'ject travels in a speci:ed direction along straight line path "rom a point A to apoint U. (n this case the distance dAU and displacement %AG will have samemagnitudes i.e d%.

Gisplacement is a vector and distance is a scalar 7uantity. An o'ject started "rom rest#travels with uni"ormacceleration.!ind the ratio o" its

displacements in # #. . . . . seconds o" its journey. & & & . . . . . . . . & 1&$&3& . . . . . . . . . & 2n1).

(" 'ody starts with an initial velocity u# and travels with uni"orm acceleration a .!ind

the ratio o" its displacements in # #. . . . . seconds o" its journey. & && . . . . . . . . & 2u8a) & 2u8$a) & 2u83a) & . . . . . . . . . & Q2u82n1)aR.

S2eed $v) & ,he distance traveled 'y a 'ody in unit time is called %peed o" the 'ody.%peedis a scalar 7uantity.

%peed v) Velocit5 $v) 6 Gisplacement o" a 'ody in unit time is called Ielocity or) -ate o" change o"displacement o" a 'ody is called its velocity. Ielocity is a vector .

Ielocity v) Wnits& Uoth speed and velocity have same units. C.S.% unit is cm / secH# !.T.% unit is

"t / secH X =.@.% or) %.( unit is m / sec.

-elation 'etween magnitude o" velocity and speed will 'e speed velocity.

!or a o'ject moving in a circular path with a constant speed# the magnitude o"velocity remains constant.Uut#the direction o" motion the o'ject continuouslychanges.

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Ielocity o" an o'ject is said to 'e constant i" a)its magnitude o" velocity is constantand ')its direction o" motion remains unchanged.

(" the velocity o" an o'ject is varying not constant) such o'ject will possess

acceleration# i.e the condition "or an o'ject to possess acceleration is v constant. An o'ject started "rom rest and traveling with uni"orm acceleration a.!ind the ratio o"

velocities at the # #. . . . . seconds o" its journey. & & & . . . . . . . . & 1&2&$& . . . . . . . . . & n.

(" 'ody starts with an initial velocity u# and travels with uni"orm acceleration a .!ind

the ratio o" its velocities at # #. . . . . seconds o" its journey. & & & . . .. . . . . & u8a) & u82a) & u8$a) & . . . . . . . . . & u8na).

Acceleration $ a ) & -ate o" change o" velocity o" a 'ody is nown as its acceleration. or)Change in velocity o" a 'ody in unit time is called its acceleration. Acceleration is a vector.

Acceleration in a 'ody may 'e due to a) change in magnitude o" velocity o" a 'odyor) ') change in direction o" velocity o" the 'ody or c) change in 'oth magnitudeand direction o" velocity o" the 'ody.

?x& !or a vertically projected 'ody or "or a "reely "alling 'ody the acceleration is dueto change in the magnitude o" velocity.

?x&!or a 'ody moving along a circular path with constant speed# the acceleration isdue to change in the direction o"velocity .

Wnits & C.S.% unit is cm/ H !.T.% unit is "t/ and =.@.% or %.( unit is m/ . Girection o" acceleration & a) !or "reely "alling 'ody the direction o" velocity and

acceleration are same# in this case acceleration is considered to 'e positive 8 ) ')!or a 'ody moving up vertically # direction o" acceleration will 'e opposite todirection velocity o" the 'ody#in this case the acceleration is considered to 'enegative ) c)!or the 'ody in circular motion the direction o" acceleration will 'eperpendicular to the direction o" velocity at every point.

Acceleration o" a 'ody may not 'e Oero# even i" the velocity o" the 'ody is Oero. ?x&!or a vertically projected 'ody # when it is at =aximum height its velocity 'ecomesOero 'ut its acceleration is e7ual to D g .

7reel5 3allin4 bod5 6 Any o'ject "alling under the in<uence o" gravitational "orce# with

acceleration due to gravity is called "reely "alling 'ody. -atio o" displacements o" a "reely "alling 'ody a"ter 1sec#2sec# . . . . . . . . . . . . . . . . n

sec o" its journey will 'e 1 & 4 & 9 & 10 . . . . . . . . . .

-atio o" displacements o" a "reely "alling 'ody 1st#2nd#$rd YY nth sec o" its journeywill 'e 1 & $ & 3 & + & 9 . . . . . . . . 2n1).

-atio o" velocity o" a "reely "alling 'ody a"ter 1 sec# 2 sec# $sec # . . . . . . . . . . . . . . o"its journey will 'e 1 & 2 & $ & 4 &

-atio o" velocity o" a "reely "alling 'ody a"ter 1st # 2nd # $rd #4th . . . . . . . . . . . . . . . . . o" its journey will 'e 1 & 1 & 1 & . . . . . . . . . . i.e its velocityin every interval is e7ual to gD.

!or a "reely "alling 'ody velocity will continuously change (ncreases) and theacceleration g ) remains constant.

Verticall5 2ro8ected /2 bod5& !or a 'ody projected vertically up with an initial velocity u#

Ielocity at maximum height will 'e Oero. At any point during its upward journey# direction o" velocity will 'e opposite to

direction o" acceleration.

Ielocity o" the o'ject at any point in its path is same in magnitude#when it is going upand coming down.

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(t reaches ground with the velocityu) with which it is projected.

Gisplacement in nth second o" its up ward journey u !or a vertically projected up 'ody velocity will continuously change decrease) and

the acceleration g ) remains constant. -atio o" velocity o" a "reely vertically projected up 'ody a"ter 1 sec# 2 sec#

$sec # . . . . . . . . . . . . . . o" its journey will 'e & & . . . . . . . . . . . . & ug) &u2g) & u$g) & . . . . . . . . . . . . . . & ung).

Pro8ectile & A 'ody projected with an angle other than Z ) is called a projectile . Tath o" a projectile is para'ola. !or a projectile as no "orce acts on it in the horiOontal direction# it does not possess

acceleration in horiOontal direction.  ,he horiOontal component o" velocity remains constant throughout its journey. gravitational "orce acts on it vertically downwards#hence it will possess acceleration

e7ual to acceleration due to gravity and vertical component o" velocity will 'edi[erent at di[erent points.

Tath o" a 'ody projected horiOontally "rom top o" a tower is a para'ola. Tath o" an o'ject dropped "rom aeroplane <ying at certain height will 'e a) para'ola

with respect to a stationary o'server# 'ut ') path o" that o'ject is a straight line withrespect to an o'server in the aeroplane.

(" a 'ody is projected horiOontally "rom the top o" a tower and another 'ody dropped"reely "rom the same height at the same time# 'oth will reach the ground at thesame time.

At maximum height o" a projectile & a)(t will possess only horiOontal velocity')6oriOontal component o" velocity c)vertical component o" velocity is

d) @.? e)T.? ") velocity is minimum# hence @.? also will'e minimum S) T.? is maximum as it is at maximum height.

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 ,agged Synopsis

 Jan 05

 9a,oo Ans0ers.1E& A particle is thrown vertically upwards with speed 2 meters per second. ,wo secondslater# another particle is thrown vertically upwards "rom exactly the same place. (ts initialvelocity is also 2 meters per second. 6ow long a"ter the :rst particle was projected# do thetwo particles collide Euestion ased 'y gloom strien in Fahoo answers).Ans & (nitial velocity o" the 1st particle 2 m /sec assume acceleration due to gravity g

1m/ .

=aximum height traveled 'y this particle .

H 2m.

7/21/2019 Problems

http://slidepdf.com/reader/full/problems-56e0cd8858d6f 18/22

 ,ime o" ascent u/g 2/1 2 sec.

i.e when the second o'ject is projected "rom ground# :rst o'ject is at maximum height .

>hen a 'ody is dropped "rom a =aximum height or tower o" height h and another 'ody is

thrown up vertically with a velocity u then they will meet a"ter th/u seconds.

t 2/21 sec

 ,hey will collide a"ter 1sec a"ter the (( particle is projected or $sec a"ter the ( particle isprojected.

Tosted in Kinematics

 ,agged Answers to your Questions.

 Jan 04

Im2ortant 3orm/las in P,5sics.6+inematics &

Average %peed (" the 'ody covers 1st hal" o" distance with a speed x and the second hal" with

a speed y#then the average speed (" the 'ody covers 1st 1/$rd o" a distance with a speed x # and 2nd 1/$ with a

speed y # and the $rd 1/$rd distance with a speed O# then average speed

Average velocity

(" a 'ody travels a displacement in seconds and a displacement in

seconds# in the same direction then Average velocity . (" a 'ody travels a displacement with velocity # and displacement with

velocity in the same direction then Average velocity (" a 'ody travels :rst hal" o" the displacement with a velocity and next hal"

o" the displacement with a velocity in the same direction # then

Average velocity .

(" a 'ody travels a time with velocity and "or a time with a velocity inthe same direction then Average velocity .

(" the 'ody travels 1st hal" o" the time with a velocity next hal" o" the time

with a velocity in same direction # then Average velocity !or a 'ody moving with uni"orm acceleration i" the velocity changes "rom u to

v in t seconds# then Average velocity u8v)/2 . E:/ations o3 motion o3 a bod5 movin4 0it, /ni3orm acceleration alon4

strai4,t line.

7/21/2019 Problems

http://slidepdf.com/reader/full/problems-56e0cd8858d6f 19/22

a) Iu8at ') %ut8 c) 2as Gistance traveled in the nth second u8an1/2) E:/ations o3 motion 3or a 3reel5 3allin4 bod5  \ote& we can o'tain these

e7uations 'y su'stitution o" u and ag in a'ove e7uations . a) vgt ')

%1/2 c) 2gs and the e7uation "or the distance traveled in nth second

changes to gn1/2) E:/ations o3 motion o3 a bod5 2ro8ected /2 verticall56we will o'tainthese e7uations 'y su'stitution ag in e7uations o" motion)

a) vugt ') %ut1/2 c) 2gs and ugn1/2)

?7uation "or maximum height reached

 ,ime o" ascent H u

 ,ime o" descent H u  ,ime o" <ight ,2u/g >hen a 'ody is thrown up "rom top o" a tower or released "rom a rising

'aloon#with velocity u.Gisplacement traveled 'e"ore reaching ground

%ut81/2 . t time during which the o'ject is in the air and %hheight o"

the tower). >hen a 'ody is dropped "rom a tower o" height h and another 'ody is thrown

up vertically with a velocity u then they will meet a"ter th/u seconds. >hen a 'ody is dropped "rom a tower o" height h . (ts velocity when it reaches

ground v

(" the displacements o" a 'ody in seconds o" itsjourney.,hen the

uni"orm acceleration o" the 'ody a !rom the a'ove e7uation we can o'serve that 'y su'stituting n1#2#$#4#Y. we

get a YYY.. a. A 'ody projected up with velocity u "rom the top o" a tower reaches ground

in seconds.(" it is thrown down with the same velocity u it reaches ground

in seconds.,hen# when it is dropped "reely the time taen to reach the groundwill 'e t and h1/2 g and 2 .

Pro8ectile motion &et us suppose that a projectile is projected with an initialvelocity u maing an angle with x axis. a)6oriOontal component o"

velocity # and #which will 'e constant through out the <ighto" the projectile as horiOontal component o" acceleration . ')Iertical

component o" velocity o" the projectile . Iertical component o"velocity at any time o" its journey gt or gt. c)=agnitude o"

the resultant velocity I and the angle x made 'y v with the

horiOontal is given 'y

 ,ime o" ascent

 ,ime o" descent

 ,ime o" <ight

=aximum height reached H when u is same

6oriOontal -ange - a)- is maximum when ')

  c) (" , is the time o" <ight# - d)!or given velocity o"

7/21/2019 Problems

http://slidepdf.com/reader/full/problems-56e0cd8858d6f 20/22

projection - is same "or the angles o" projections and ?x& 23 and 03

i.e the -ange "or those two angles will 'e same whose sum is )

Tosted in Kinematics

 ,agged ormula!s

"ec 31

De1nitions -+inematics+inematics : The study of motion of bodies, does not taking in to consideration the cause

for motion is called Kinematics.

est: If the position of an object remains to be same for any length of time with respect to

its surroundings, then the object will be at rest with respect to those surroundings.

Motion& If the position of a body is continuously changing with respect its surroundings

with time , then the body is said to be in motion.

"ni3orm motion & (" a 'ody moving along a straight line path travels e7ual distances in

e7ual interval o" time# then the 'ody is said to 'e in uni"orm motion.

Non "ni3orm motion & (" a 'ody moving along straight line travels une7ual distances in

e7ual intervals o" time# the 'ody is said to 'e in \on uni"orm motion. or) (" a 'ody has

une7ual displacements in in e7ual intervals o" time# it will 'e in \on uni"orm motion.

Dis2lacement $ S ) : The change in the position of a body in a specific direction, is called

Displacement. Displacement is a vector uantity.

S2eed $v) : The distance traveled by a body in unit time is called !peed of the body.%peed

is a scalar 7uantity.

%peed v) .

"ni3orm s2eed & (" the distance traveled 'y a 'ody in e7ual intervals o" time are same# "orany length o" time then the 'ody is said to 'e in uni"orm speed.

Non-/ni3orm s2eed & (" the distance traveled 'y a 'ody in e7ual intervals o" time are

di[erent#the 'ody is said to 'e in non uni"orm speed.

Velocit5 $v) 6 Gisplacement o" a 'ody in unit time is called Ielocity or) -ate o" change o"

displacement o" a 'ody is called its velocity. Ielocity is a vector .

Ielocity v) .

"ni3orm velocit5 $v) & (" the displacements o" a 'ody in e7ual intervals o" time are same#

"or any length o" time then the 'ody is said to 'e in uni"orm motion.

Non-/ni3orm velocit5 & (" the displacements o" a 'ody in e7ual intervals o" time are

di[erent#the 'ody is said to 'e in non uni"orm motion.

Instantaneo/s Velocit5& ,he velocity o" a 'ody at a particular instant o" time is nown as

instantaneous velocity.

Avera4e Velocit5 & (" the displacement o" a 'ody in a small interval o" time dt is ds then#

the ratio o" the displacement ds to the time interval dt is called average velocity. Average

velocity v ds)/dt.

7/21/2019 Problems

http://slidepdf.com/reader/full/problems-56e0cd8858d6f 21/22

Initial velocit5 $ / ) & Ielocity o" the 'ody in the 'eginning o" the o'servation is called

initial velocity.

7inal velocit5 $ v ) & Ielocity o" the 'ody at the end o" the o'servation is called its :nal

velocity.

Acceleration $ a ) & -ate o" change o" velocity o" a 'ody is nown as its acceleration. or)

Change in velocity o" a 'ody in unit time is called its acceleration. Acceleration is a vector.

Deceleration & (" the velocity o" a 'ody is continuously decreasing with time# the 'ody will

possess negative acceleration. (t is also called deceleration.

"ni3orm acceleration& (" the change increase or decrease ) in velocity o" a 'ody is

constant in e7ual intervals o" time# the 'ody will possess uni"orm acceleration.

Non /ni3orm acceleration & (" the change increase or decrease ) in velocity o" a 'ody is

di[erent in e7ual intervals o" time# the 'ody will possess nonuni"orm acceleration.

Acceleration d/e to 4ravit5 $ 4 )6 The acceleration in a freely falling body due togravitational force acting on it is called acceleration due to gravity. This is a vector. The

value of g in ".#.! system is $%& and in '.k.s ( !.I systems is $.% .

7reel5 3allin4 bod5 6 Any o'ject "alling under the in<uence o" gravitational "orce# with

acceleration due to gravity is called "reely "alling 'ody.

Ma;im/m ei4,t $ ) : The distance traveled by a body vertically projected up, just

before coming to rest is called, ma)imum height.

Time o3 ascent  ) & ,he time taen 'y a vertically projected 'ody to reach the maximum

height is called time o" ascent.

Time o3 descent  ) & ,he time taen 'y a 'ody to reach the ground "rom its maximum

height is called time o" descent.

Time o3 <i4,t * T + :The total time during which the vertically projected body will remain in

air is called time of flight.

* or + Total time taken by a projectile to reach the same horiontal plane from which it is

projected is called called time of flight.

 ,ime o" <ight , ) 8 ) .

Pro8ectile : - body projected with an angle other than * + is called a projectile .

ori=ontal an4e * / + :'a)imum horiontal distance traveled by the body before it

touches the ground *or+ 'a)imum horiontal distance traveled by the body before it touches

the point on the same level of projection is called horiontal range * / +.

Challenge Problem

 

7/21/2019 Problems

http://slidepdf.com/reader/full/problems-56e0cd8858d6f 22/22

10) Seth is doing his student driving with the “Give-Me-A-Brake-PLEASE” driving

schoo and is traveing down the interstate with a s!eed o" #$0%&s$ Mack is driving his

“1'-wheeer” down the "ast ane at ($0%&s when he notices Seth *0$0%eters ahead o"

hi% in the right ane$

a)  +" Mack and Seth %aintain their s!eeds, how "ar %ust Mack trave e"ore he

catches u! to Seth. )  /ow ong wi this take.