PDT 180ENGINEERING SCIENCEVectors And ScalarsVectors And Scalars(Continue)

PROJECTILE MOTION

A projectile is an object moving in two dimensions under the influence of Earth's gravity.

Its path is a parabola.

• Neglect air resistance.

• Consider motion only after release and before it hits.

• Analyze the vertical and horizontal components separately (Galileo).

• No acceleration in the horizontal, so velocity is constant.

• Acceleration in the vertical is – 9.8 m/s2 due to gravity and thus velocity is not constant.

• Object projected horizontally will reach the ground at the same time as one dropped vertically.

EQUATIONS FOR PROJECTILE MOTION

00 xvv

tvxx x00

tgvv yy 0

)(2 02

02 yygvv yy

200 2

1tgtvyy y

Horizontal Vertical

ax=0 ay = - g

vx= constant

INITIAL VELOCITY

cos00 vvx

•If the ball returns to the y = 0 point, then the velocity at that point will equal the initial velocity.

•At the highest point, v0 y = 0 and v = vx0

sin00 vvy

Module 7 - 5

EXAMPLE 3A

)0.50)(cos0.18(0

smv

x s

m6.11

)0.50)(sin0.18(0

smv

y

A football is kicked at an angle of 50.00 above the horizontal with a velocity of 18.0 m / s. Calculate the maximum height. Assume that the ball was kicked at ground level and lands at ground level.

sm8.13

tgvv yy 0 0

g

vt y 0

280.9

8.13

sms

m s41.1

2

0max 2

1gttvyy

yo

22max )41.1)(8.9(

2

1)41.1)(8.13(0 s

smss

my

my 7.9max

at top:

Module 7 - 9

EXAMPLE 4A

)0.50)(cos0.18(0

smv

x s

m6.11

)0.50)(sin0.18(0

smv

y

A football is kicked at an angle of 50.00 above the horizontal with a velocity of 18.0 m / s. The football hits a window in a house that is 25.0 m from where it was kicked. How high was the window above the ground.

sm8.13

tvxx 0

0xv

xt

smm

6.11

0.25 s16.2

2

00 2

1gttvyy

y

22 )16.2)(8.9(

2

1)16.2)(8.13(0 s

smss

my

my 9.6

Time to hit the window:

Module 7 - 10

EXAMPLE 4 B

st 16.2gtvv

yy

0

)16.2()8.9()8.13( 2 ss

ms

mvy

What is the final velocity and angle of the football that hit the window in Example 4 A.

sm37.7

smv

x6.11

22 )37.7()6.11( sm

smv

sm7.13

x

y

v

vtan

6.11

37.7tan 1 4.32 below x axis

Module 7 - 12

Example 5. (35) A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. If the plane is traveling horizontally with a speed of 250 km /h (69.4 m / s) how far in advance of the recipients (horizontal distance) must the goods be dropped (Fig. 3–37a)? .

smv

v

x

y

/4.69

0

0

0

Coordinate system is 235 m below plane

02

10235 2 tgmy mx

ssmtvxx xo

481

)93.6()/4.69(00

ssm

mt 93.6

/8.9

)235()2(2

Module 7 - 13

Projectile Motion Is ParabolicIn order to demonstrate that projectile motion is parabolic, the book derives y as a function of x. When we do, we find that it has the form:

This is the equation for a parabola.