Julie throws a ball to her friend Sarah. The ball leaves Julie\'s hand a distanc
ID: 1593557 • Letter: J
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 24 m/s at an angle 35 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?
m/s
2)
What is the vertical component of the ball’s velocity when it leaves Julie's hand?
m/s
3)
What is the maximum height the ball goes above the ground?
m
4)
What is the distance between the two girls?
m
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 23 m/s when it reaches a maximum height of 13 m above the ground.
What is the speed of the ball when it leaves Sarah's hand?
m/s
6)
How high above the ground will the ball be when it gets to Julie? (note, the ball may go over Julie's head.)
m
Explanation / Answer
(a)
Initial Horizontal Speed, Vx = 24 * cos(35) m/s = 19.66 m/s
Horizontal Acceleration, a = 0
Horizontal component of the ball’s velocity when it leaves Julie's hand, Vx = 19.66 m/s
(b)
Initial Vertical Component, = 24 * sin(35) m/s = 13.77 m/s
Vertical component of the ball’s velocity when it leaves Julie's hand, Vy = 13.77 m/s
(c)
Vertical Acceleration = 9.8 m/s^2 (downwards)
At Max Height Vertical component of velocity = 0
V^2 = u^2 - 2*a*s
0 = 13.77^2 - 2*9.8*s
S = 9.67 m
Maximum height the ball goes above the ground, = 9.67 + 1.5
Maximum height the ball goes above the ground, = 11.17 m
d)
Calculating time ball is in air,
Using Newton law of motion,
S = u*t - 1/2*at^2
0 = 13.77*t - 1/2 * 9.8 * t^2
t = 2.81 s
Distance between the two girls, = Vx*t
S = 19.66 * 2.81 m
S = 55.24 m
Distance between the two girls, S = 55.24 m
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