Question

An object of mass m is projected with a momentum $ρ$ at such an angle that its maximum height is 1/4th of
its horizontal range. Its minimum kinetic energy in its path will be

Moderate

Solution

It is given that, $H = \frac{R}{4}$
$∴ \frac{u^{2} \sin^{2} θ}{2 g} = \frac{2 u^{2} \sin θ \cos θ}{4 g} or \tan θ = 1 \Rightarrow θ = 4$ 5⁰
At highest point, momentum will remain $\frac{p}{\sqrt{2}}$
$∴ K = \frac{(p / \sqrt{2})^{2}}{2 m} = \frac{p^{2}}{4 m}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME