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 5.22 m/s. The train has a velocity of 24 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

(a) v = u + a t
t = ( v - u ) / a ... where v = 5.22; u = 0; a = -9.8
to the ball's flight apex ;
t = ( 5.22 - 0 ) / 9.8 0.533 s;

..... so the total time in flight = 2 x 0.533 s 1.07 s
assuming the ball is caught at the same height as it was released.

(b) observed velocity = ( 24 + 5.22 i ) m/s = (24.56 12.3°) m/s
speed = mod(24.56 12.3°) m/s 24.56 m/s ; and
launch angle = arg(24.56 12.3°) m/s 12.3°
.......................................... ( 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 = 24 * 0.533 12.79 m
vertical component of displacement = Sy = u t + 0.5 a t ²
Sy = 5.22*0.533 + 0.5 *( 9.8 )* 0.533 ²
= 2.78 - 1.39 1.39 m
total displacement = mod( 12.79 + 1.39 i ) 12.87 m

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