Answer to Question #235989 in Classical Mechanics for AJS

Question

A juggler traveling in a train on level track throws a ball
straight up, relative to the train, with a speed of 4.90 m/s. The train
has a velocity of 20.0 m/s due east. As observed by the juggler,
(a) what is the ball’s total time of flight, and (b) what is the displacement of the ball during its rise? According to a friend standing
on the ground next to the tracks, (c) what is the ball’s initial speed,
(d) what is the angle of the launch, and (e) what is the displacement
of the ball during its rise?

Answer 1

(a) The time is

$$T=2\frac {v_b} g=1\text{ s}.$$

(b) The displacement is

$$d_b=\frac{v_b^2}{2g}=1.23\text{ m}.$$

(c) The initial speed is

$$v=\sqrt{v_b^2+v_t^2}=20.6\text{ m/s}.$$

(d) The angle is

$$\theta=\arctan\frac{v_b}{v_t}=13.8°.$$

(e) The displacement is

$$d=\sqrt{d_b^2+d_x^2}=\sqrt{d_b^2+\bigg(v_t\frac{T}{2}\bigg)^2}=10.1\text{ m}.$$