A boy on a train of height `h_1` projects a coin 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

(a) `h_2 - h_1 = x tan theta - (gx^2)/(2 v^2 cos^2 theta)` from here, we will get two values of `x`, both can be valid ;
(b) `H = h_1 + (v^2 sin^2 theta)/(2g)` ;
( c) `sqrt(v^2 - 2g(h_2 - h_1))`.

Explanation

(a) Taking point `A` as origin
`y = x tan theta - (gx^2)/(2 u^2 cos^2 theta)`
`rArr h_2 - h_1 = x tan theta - (gx^2)/(2 y^2 cos^2 theta)`
From here, we will get two values of `x`, both can be valid.
(b) Maximum height `H = h_1 + (v^2 sin^2 theta)/(2 g)`
( 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' + 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)`.
.