Question

Find the distance along the bottom of the box between the point of projection P and the point Q where the particle lands. (Assume that the particle does not hit any other surface of the box. Neglect air resistance)

Answer 1

Assume $x$ is the direction along the plane and $y$ is the direction perpendicular to the plane.
Acceleration of particle w.r.t block $=$ Acceleration of particle - Acceleration of block
$=(gsin\theta \hat{i}+gcos\theta \hat{j})gsin\theta \hat{j}=gcos\theta \hat{j}$
$a_x=0$
$a_y=-gcos\theta$; for upward movement
In $y$ direction:

$h=0=\left(usin\theta \right)t\left(gcos\theta \right)t^2$

$t=\frac{2usin\theta }{gcos\theta }$
Along the horizontal surface:
$s=ucos\theta \times t=\frac{u^{2}sin2\theta }{gcos\theta }$
Along the bottom of the box:
$s=ucos\theta \times t\times \frac{1}{cos\theta }=\frac{u^{2}sin2\theta }{(gcos\theta {)}^2}$