Topics for Test #2 Vectors/Projectiles

1.Vector addition using head to tail2.Resultant vectors from components Vx an Vy

3.Calculating an angle from given vectors4.Utilizing sin, cos, tan, & pythagorean theorem to solve triangles5.Solve horizontally launched projectiles for time, vx , Dx, or Dy6.Solve ground launched projectiles for Range, height, or muzzle velocity7.Understand the impact velocity of a projectile8.Calculate the x and y components of a projectiles muzzle velocity9.Relate information from Lab #4 Free Fall, Lab #5 Nerf Gun, and Computer Simulation (Canon)

Equations will be provided, but you need to be familiar with them!

Projectiles launched from aboveHorizontal : vx = dx / t Dx = vxt

Vertical : vy = -gt dy = vit + ½ gt2

Projectiles launched from ground @ angleHorizontal : R = vi2 Sin 2 (q) / g

Vertical : h = vi2 Sin2 (q) /2 g

Total Time : T = 2vy /g

1.What is the net force acting in the x dimension for the following car?

Fengine Fwind Ffriction

1,000N 100N -500N

Calculate the missing value

100 m/s

Vx

53.1°

3. A soccer ball is kicked at ground level with a velocity of 27.0m/s at an angle of 30 degrees.a)what is the hang time in the air?b)What is the maximum height?c) What is the range?

3.A dart is thrown straight up in the air with a velcoity of 90 m/s.

a) How much time will it take to go up?

b) How much time wil it take to come back down?

c)Total time in the air is?

d)What height will it reach?

A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high.

a) How long does it take for the stone to reach the bottom of the cliff?

b) How far from the base of the cliff does the stone hit the ground?

5 m/s

digram?

30 m/s

5.A boat is traveling east at 100 m/s. A current flows

south at 20 m/s. The river is 1,000m across.

a) Find the resulting vector w/ direction

b) How much time does it take to get across the river?

c) What is the displacement from A to B on the eastern shore line?

A

B

Answer key1.

2.

1,000 100 -500 600

a) vy = sin (30) 27 m/s = 13.5 m/svy =- gt so -13.5m/s = 9.8(t) = 1.4 sec

then 1.4 s x 2 = 2.8 s

or Total time = 2Vy / g = 2.8 secs

b) Dy= 1/2 gt2= (.5) 9.8 (1.4)2

= 37m

c) sin (60) 27m/s = 23.4 m/si ) Time= 4.77sii) y =111.6 m/s

3. a) Vy =-gt b) 9.2 secs c) 9.2 x 2 = 18.4 secs d) y= 1/2 gt2 -90 m/s =- 9.8 t y= 413m

t= 9.2 secs

30 m/s 30 deg

400 m/s

150 deg*The large angle is 150 deg from geometry ( 30+150=180)

C2 = a2+b2+2abcos(q)C2 = 302+4002+2(30)(400)cos(150 )C= 413 m/s

4.

5. 100 m/s

20 m/s

a) v2= (100)2+(20)2

v= 101.9 m/s SE

b) River is 1,000m in dx

dx = vx t1,000 = (100 m/s) t

t= 10 secs

c) dx = 20 m/s (10s)dx= (100 m/s) tt=200 m

This is like a projectile problem!