Answer
a). The sketch is as follows.
b). $t=2.55s$
c). From the viewpoint of the person in railcar, the velocity of the ball at the top will be zero. However, viewing from outside the moving train, the ball will maintain its initial horizontal velocity component. Viewed from the person next to the tracks is 12m/s on the direction of the train movement.
d). $\theta=64^{\circ}$
e) $x=30.6m$
Work Step by Step
a). The sketch is as attached.
b). $t=\frac{v-u}{g}=\frac{25}{9.8}=2.55s$
c). From the viewpoint of the person in railcar, the velocity of the ball at the top will be zero. However, viewing from outside the moving train, the ball will maintain its initial horizontal velocity component. Viewed from the person next to the tracks is 12m/s on the direction of the train movement.
d). $\theta=tan^{-1}\frac{25}{12}=64^{\circ}$
e). $x=velocity \times time=12\times 2.55=30.6m$
