Question

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 18 m/s at an angle 42 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.
1) What is the horizontal component of the ball’s velocity when it leaves Julie's hand?
2) What is the vertical component of the ball’s velocity when it leaves Julie's hand?
3) What is the maximum height the ball goes above the ground?
4) What is the distance between the two girls?
5) After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah's hand a distance 1.5 meters above the ground, and is moving with a speed of 14 m/s when it reaches a maximum height of 10 m above the ground.
What is the speed of the ball when it leaves Sarah's hand?
6) How high above the ground will the ball be when it gets to Julie? (note, the ball may go over Julie's head.)

Answer 1

Answer:

1) 13.377 m/s

2) 12.044 m/s

3) 8.893 m

4) 32.85 m

5) 19.05 m/s

6) 3.25 m

Explanation:

1)

V_o,x = V_o * cos (Q)

V_o,x = 18 * cos (42)

V_o,x = 13.377 m/s

2)

V_o,y = V_o * sin (Q)

V_o,y = 18 * sin (42)

V_o,y = 12.044 m/s

3)

Maximum height is reached when V,y = 0

V,y = V_o,y + a*t

0 = 12.044 - 9.81t

Solve above equation for t:

t = 1.228 s

Compute S_y @t = 1.228 s

S_y = S_o,y + V_o,y*t + 0.5*a*t^2

S_y = 1.5 + 12.044*1.228 - 4.905*1.228^2

S_y = 8.893 m

4)

Time taken for the ball to complete path:

S_y = S_o,y + V_o,y*t + 0.5*a*t^2 = 1.5

V_o,y*t + 0.5*a*t^2 = 0

12.044*t - 4.905*t^2 = 0

t = 0, t = 2.455 s

Total distance traveled in horizontal direction S_x @ t = 2.455 s

S_x = S_o,x + V_o,x*t

S_x = 0 + 13.377*2.455 = 32.85 m

5)

S_y = S_o,y + V_o,y*t + 0.5*a*t^2 = 10

10 = 1.5 + V_o,y*t -4.905*t^2   .... Eq 1

Maximum height is reached when V,y = 0

V,y = V_o,y + a*t

0 =  V_o,y - 9.81t  .... Eq2

Solve Eq 1 and Eq 2 simultaneously

V_o,y = 9.81*t

10 = 1.5 + 9.81*t^2 -4.905*t^2

8.5 = 4.905*t^2

t = 1.316 s

V_o,y = 12.914 m/s

Compute Velocity

V = sqrt (V_o,x^2 + V_o,y^2)

V = sqrt (14^2 + 12.914^2)

V = 19.05 m/s

6)

Total distance traveled in horizontal direction between players is 32.85m

S_x = S_o,x + V_o,x*t

S_x = 0 + 14*t = 32.85 m

t = 2.3464 s

Compute Sy @ t = 2.3464 s

S_y = S_o,y + V_o,y*t + 0.5*a*t^2

S_y = 10  - 4.905*(1.1732)^2

S_y = 3.25 m