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 5.15 m/s. The train has a velocity of 20 m/s due east. As observed by the juggler, what is the ball's total time of flight? What is the displacement of the ball during its rise? 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 What is the displacement of the ball during its rise? x-component: m y-component: m

Explanation / Answer

(a) v = u + a t
t = ( v - u ) / a ... where v = 5.15; u = 0; a = -9.8 m/s^2
to the ball's flight apex ;
t = ( 5.15 - 0 ) / 9.8 0.526 s;
..... so the total time in flight = 2 x 0.526 s 1.052 s
assuming the ball is caught at the same height as it was released.

(b) observed velocity = ( 20 + 5.15 i ) m/s = (20.65 14.4°) m/s
speed = mod(20.65 14.4°) m/s 20.65 m/s ; and
launch angle = arg(20.65 14.4°) m/s 14.4°
.......................................... ( elevation relative to the horizontal ) .

(c) Not sure which one you mean here:
........straight line distance between launch point and apex ; or
........exact path length travelled by ball along parabolic arc.
so I will do the easier one : ()
horizontal component of displacement = Vx t = 20 * 0.526 10.52 m
vertical component of displacement = Sy = u t + 0.5 a t ²
Sy = 5.15*0.526 + 0.5 *( 9.8 )* 0.526 ²
= 2.7 - 1.36 1.34 m

total displacement = mod( 10.52 + 1.34 i ) 10.6 m

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