Question

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

• A. $\frac{u^{2}sin2\alpha }{gcos\theta }$
• B. $\frac{u^{2}cos2\alpha }{gcos\theta }$
• C. $\frac{u^{2}sin2\alpha }{2gcos\theta }$
• D. $\frac{u^{2}sin2\alpha }{2gsin\theta }$

Answer 1

The correct option is A $\frac{u^{2}sin2\alpha }{gcos\theta }$

Take the x-axis along incline and y-axis perpendicular to the plane.

$\therefore$ Acceleration of particle = gsin$\theta \hat{i}$+ gcos$\theta \hat{j}$and acceleration of block = gsin$\theta \hat{i}$

$\therefore$ Acceleration of particle with respect to block = acceleration of particle - acceleration of block= (gsin$\theta \hat{i}$+ gcos$\theta \hat{j}$) -(gsin$\theta \hat{i}$) = gcos$\theta \hat{j}$

Now motion of particle with respect to block will be a projectile as shown.

The only difference is, g will be replaced by g cos$\theta$

PQ= Range(R)= $\frac{u^{2}sin2\alpha }{gcos\theta }$

PQ= $\frac{u^{2}sin2\alpha }{gcos\theta }$