Julie throws a ball to her friend Sarah. The ball leaves Julie\'s hand a distanc

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 22 m/s at an angle 62 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 11 m/s when it reaches a maximum height of 22 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.)

(Survey Question) 7) Below is some space to write notes on this problem

Explanation / Answer

1) vh = 22*cos(62) = 10.32 m/s

2) vv = 22*sin(62) = 19.42 m/s

3) Hmax = u^2*sin^2(theta)/(2*g) = 22^2*sin^2(62)/(2*9.81) = 19.23 m

4) Range R = u^2*sin(2*theta)/g = (22^2*sin(2*62))/9.81 = 40.9 m

5) at maximum height u*cos(theta) = 11

and u^2*sin^2(theta)/(2*g) = 22

u^2*sin^(theta) = 22*2*9.81 = 431.64

u*sin(theta) = 20.77

u*cos(theta) = 11

tan(theta) = 20.77/11

theta = tan^(-1)(1.88) = 62 degrees

u = 11/cos(62) = 23.43 m/s

6) horizontal distance travelled is R = u*cos(theta)*t

40.9 = 23.43*cos(62)*t

t = (40.9)/(23.43*cos(62))

t = 3.71 s

y = (u*sin(62)*t)-(0.5*g*t^2)

y = (23.43*sin(62)*3.71)-(0.5*9.81*3.71^2) = 9.23 m

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