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 16 m/s at an angle 32 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 right before Sarah catches it? (13.56 m/s)

2)

What is the vertical component of the ball’s velocity right before Sarah catches it? (-8.48m/s)

3)

What is the time the ball is in the air?(1.73s)

4)

What is the distance between the two girls?(23.48m)

5)

After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it reaches Julie's horizontal position. Assume the ball leaves Sarah's hand a distance 1.5 meters above the ground, reaches a maximum height of 8 m above the ground, and takes 1.466 s to get directly over Julie's head.

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?

# 5 & 6 i do not know how to solve them?

Explanation / Answer

here,

initial velocity of the ball , u = 16 m/s

angle ,theta = 32 degree

1)

the horizontal velocity , ux = 16* cos(32)

ux = 13.56 m/s

as there is no horizontal accelration

the horizontal component of the ball’s velocity right before Sarah catches it is 13.56 m/s

2)

the vertical velocity , uy = 16* sin(32)

uy = 8.48 m/s

the velocity is ball is in downward direction , when tit reahes sarah

therefore ,

the vertical component of the ball’s velocity right before Sarah catches it is - 8.48 m/s

3)

the time the ball is in air be t

using seccond equation of motion

0 = 8.48 * t - 0.5 * 9.8 * t^2

t = 1.73 s

the ball is in the air for 1.73 s

4)

the distance between the girld , d = ux * t

d = 13.56 * 1.73

the distance between the girls is 23.48 m

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