Question

A particle is projected from the bottom of an inclined plane of angle ${30}^{\circ }$ with a velocity of $30\text{m/s}$ at an angle of ${60}^{\circ }$ with the horizontal. Find the time of flight.
(Take $g=10m/s^2$)

• A. $\sqrt{3}\text{s}$
• B. $\frac{1}{\sqrt{3}}\text{s}$
• C. $2\sqrt{3}\text{s}$
• D. $2\text{s}$

Answer 1

The correct option is C $2\sqrt{3}\text{s}$

We know that time of flight of a projectile on an inclined plane is given by
$T=\frac{2u\sin (\alpha -\theta )}{g\cos \theta }$
Here $\alpha ={60}^{\circ }$ and $\theta ={30}^{\circ }$
$\therefore T=\frac{2\times 30\times \sin ({60}^{\circ }-{30}^{\circ })}{g\cos {30}^{\circ }}$
$\Rightarrow T=2\sqrt{3}\text{s}$