Question

A large heavy box is sliding without friction down a smooth plane of inclination $\theta$. From a point $P$ on the bottom of the box, a particle is projected inside the box. The initial speed of particle with respect to the box is $u$ and the direction of projection makes an angle $\alpha$ with the bottom as shown. Find the distance along the bottom of box between the point of projection $P$ and point $Q$ where the particle lands. (Assume that the particle does not hit any other surface of the box. Neglect air resistance)

• A. $\frac{u^2\sin 2\alpha }{2g}$
• B. $\frac{u^2{\sin }^2\alpha }{2g\cos \theta }$
• C. $\frac{u^2\sin 2\alpha }{g\cos \theta }$
• D. $\frac{u^2\sin 2\alpha }{g}$

Answer 1

The correct option is C $\frac{u^2\sin 2\alpha }{g\cos \theta }$

Considering the box as the frame of reference,
For the projectile ,
Initial velocity $=u$
Inclination $=\alpha$
$\therefore$ Range $=\frac{u^{2}sin2\alpha }{g^{\prime }}$
Here,
$g^{\prime }$ is component of acceleration in the vertical plane of box.
i.e. $g^{\prime }=gcos\theta$
$\therefore$ Range $=\frac{u^{2}sin2\alpha }{gcos\theta }$