Question

A particle is projected up the incline such that its component of velocity along the incline is $10\text{m/s}$. Time of flight is $2\text{sec}$ and maximum height above the incline is $5\text{m}$. Then velocity of projection will be

• A. $10\text{m/s}$
• B. $10\sqrt{2}\text{m/s}$
• C. $5\sqrt{5}\text{m/s}$
• D. $5\sqrt{10}\text{m/s}$

Answer 1

The correct option is B $10\sqrt{2}\text{m/s}$


Let the direction along the inclined plane be $x$ and the direction perpendicular to the inclined plane be $y$.
Let $T$ be the time of flight of the projectile.
Using third equation of motion along $y$-axis, we get
$v_y^2=u_y^2-2g\cos \theta h$
where $h$ is the height above the inclined plane.

At maximum height, $v_y=0$
$\Rightarrow 0=u^2{\sin }^2(\theta -a)-2g\cos a\times 5\Rightarrow u^2{\sin }^2(\theta -a)=10g\cos a...\left(i\right)$

We know that time of flight of a projectile on an inclined plane is given by
$T=\frac{2u\sin (\theta -a)}{g\cos \alpha }$
$\Rightarrow 2=\frac{2u\sin (\theta -\alpha )}{g\cos \alpha }\Rightarrow u\sin (\theta -\alpha )=g\cos \alpha ...\left(ii\right)$

Dividing (i) from (ii), we get
$u\sin (\theta -\alpha )=10$
We already know $u\cos (\theta -\alpha )=10$
Squaring and adding,
$u^2{\sin }^2(\theta -a)+u^2{\cos }^2(\theta -a)=200\Rightarrow u=10\sqrt{2}{\text{ms}}^{-1}$