Question

The potential energy of a projectile at its maximum height is equal to its kinetic energy there. If velocity of projection is $20m{s}^{-1}$, its time of flight is $(g=10m{s}^{-2})$

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

Answer 1

The correct option is B $2\sqrt{2}s$

$\text{Step 1: Represent the projectile [Refer Fig.]}$

Let $\theta$ be the angle of projection with the horizontal.

Let $u$ be the velocity of the projectile.

$\text{Step 2: Finding the angle}\theta$

Maximum height of projectile: $h=\frac{u^2{\sin }^2\theta }{2g}$

At maximum height, $v=u_x=ucos\theta$

Now at maximum height :

$KE=PE$ (Given)

$\Rightarrow \frac{1}{2}m{v}^2=mgh$

$\Rightarrow \frac{1}{2}m(u\cos \theta {)}^2=mg\left(\frac{u^2{\sin }^2\theta }{2g}\right)$

$\Rightarrow {\tan }^2\theta =1$

$\Rightarrow \theta ={45}^{\circ }$

$\text{Step 3: Time of flight calculation}$

$T=\frac{2u\sin \theta }{g}$

$=\frac{2\times 20\sin {45}^o}{10}$

$\Rightarrow T=2\sqrt{2}s$

Hence option $B$ is correct