A juggler traveling in a train on level track throws a ball straight up, relativ

Question

A juggler traveling in a train on level track throws a ball straight up, relative to the train, with a speed of 5.03 m/s. The train has a velocity of 22 m/s due east. (a) As observed by the juggler, what is the ball's total time of flight? s What is the displacement of the ball during its rise? m (b) According to a friend standing on the ground next to the track, what is the ball's initial speed? m/s What is the angle of the launch? ° above the horizontal (c) What is the displacement of the ball during its rise? x-component: m y-component: m

Explanation / Answer

v = 5.03 j^ m/s
vt = 22 i^ m/s

Time taken to reach top, = t
vf = v - a*t
0 = 5.03 - 9.8 * t
t = 0.513 s
As observed by the juggler, ball's total time of flight, = 2*t = 2*0.513 = 1.03 s

(b)
Initial speed of ball, v = sqrt(22^2 + 5.03^2)
v = 22.57 m/s

Angle of launch, = tan^-1(5.03/22)
Angle of launch, = 12.88o above the Horizontal

(c)
During it's Rise Displacement of the ball
X component = 22 * 0.513 = 11.29 m

Y component , s = u*t - 1/2*a*t^2
s = 5.03 * 0.513 - 1/2 * 9.8 * 0.513^2
s = 1.29 m

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