Question

An object of mass $m$ is projected with a momentum $P$ at such an angle that its maximum height $\left(H\right)$ is $\left(\frac{1}{4}\right)th$ of its horizontal range $\left(R\right)$. The minimum value of its kinetic energy in the path will be:

• A. $\frac{P^2}{8m}$
• B. $\frac{P^2}{4m}$
• C. $\frac{3P^2}{4m}$
• D. $\frac{P^2}{m}$

Answer 1

The correct option is B $\frac{P^2}{4m}$

We have $h=\frac{v^{2}si{n}^2\theta }{2g}=\frac{1}{4}R=\frac{v^{2}sin2\theta }{4g}$

Solving this equation we get
$\theta ={45}^0$.
Thus the x and y components of the initial velocity are $\frac{v}{\sqrt{2}}$ each.
At the maximum height the y component of the velocity is zero and thus is the point of minimum kinetic energy.
Thus we get the minimum kinetic energy as

$\frac{1}{2}m{\left(\frac{v}{\sqrt{2}}\right)}^2=\frac{P^2}{4m}$