Question

a train moves to the right at speed 20 m/s relative to the ground. a passenger standing inside the train throws a ball straight up with initial vertical speed 5 m/s measured in the train frame, from a height 1.6m above the floor of the train. (a) what is the balls horizontal velocity relative to the ground immediately after the throw? (b) how long is the ball in the air before it returns to the same height/ the throwers hand? (c) how far horizontally (relative to the ground) does the ball travel while in the air?

Explanation:

Step1: Find the initial velocity relative to the ground

The train is moving horizontally at $v_{x,train}=20$ m/s and the ball is thrown vertically with $v_{y,0}=5$ m/s in the train - frame. The initial velocity relative to the ground has horizontal and vertical components. The horizontal component of the ball's velocity relative to the ground $v_{x,0}=20$ m/s and the vertical component $v_{y,0}=5$ m/s. The magnitude of the initial velocity relative to the ground is $v_0=\sqrt{v_{x,0}^2 + v_{y,0}^2}=\sqrt{20^2 + 5^2}=\sqrt{400 + 25}=\sqrt{425}\approx20.6$ m/s. The direction $\theta=\tan^{- 1}(\frac{v_{y,0}}{v_{x,0}})=\tan^{-1}(\frac{5}{20})=\tan^{-1}(0.25)\approx14.0^{\circ}$ above the horizontal.

Step2: Find the time - of - flight

We use the vertical motion of the ball. The vertical displacement $\Delta y = 0$ (returns to the same height), and the equation for vertical displacement is $\Delta y=v_{y,0}t-\frac{1}{2}gt^2$, where $g = 9.8$ m/s². Setting $\Delta y = 0$, we get $0=v_{y,0}t-\frac{1}{2}gt^2=t(v_{y,0}-\frac{1}{2}gt)$. One solution is $t = 0$ (corresponds to the initial time). The non - zero solution is $t=\frac{2v_{y,0}}{g}=\frac{2\times5}{9.8}\approx1.02$ s.

Step3: Find the horizontal distance

The horizontal motion of the ball is a uniform - motion with constant horizontal velocity $v_{x}=20$ m/s. Using the formula $x = v_{x}t$, with $t = 1.02$ s, we get $x=20\times1.02 = 20.4$ m.

Answer:

(a) The magnitude of the initial velocity relative to the ground is approximately $20.6$ m/s at an angle of approximately $14.0^{\circ}$ above the horizontal.
(b) The time the ball is in the air is approximately $1.02$ s.
(c) The horizontal distance the ball travels relative to the ground is approximately $20.4$ m.