Question

A boy on a train of height $h_1$ projects a coin to his friend of height $h_2$ standing on the same train, with a velocity $v$ relative to the train, at an angle $\theta$ with horizontal. If the train moves with a constant velocity $v$ in the direction of x-motion of the coin, find the
(a)distance between the boys so that the second boy can catch the coin,
(b) maximum height attained by the coin, and
(c) speed with which the second boy catches the coin relative to himself (train) and ground.

Answer 1

a.)Taking point $A$ as origin

$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

$\Rightarrow h_2-h_1=x\tan \theta -\frac{g{x}^2}{2v^2{\cos }^2\theta }$

From here, we will get two value of $c$, both can be valid.

b.) Maximum height $H=h_1+\frac{v^2{\sin }^2}{2g}$

c.) w.r.t. train:

$v_x=v\cos \theta$

$v_y^2=(v\sin \theta {)}^2-2g(h_2-h_1)$

final speed :

$v_f=\sqrt{v_x^2+v_y^2}=\sqrt{(v\cos \theta {)}^2+(v\sin \theta {)}^2-2g(h_2-h_1)}$

$=\sqrt{v^2-2g(h_2-h_1)}$

w.r.t ground $v_x=v^{\prime }+v\cos \theta$

$v_y^2=(v\sin \theta {)}^2-2g(h_2-h_1)$

Final Speed $v_f=\sqrt{v_x^2+v_y^2}4$