Question

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

Answer 1

Given that,

Maximum height of the projectile is equal to 1/4th of its horizontal range.

To find,

The minimum kinetic energy in its path will be?

Solution,

Let H is maximum height and R is horizontal range. ATQ,

$H=\dfrac{R}{4}$

Putting the formula for H and R for projectile motion. So,

$\dfrac{u^2\sin^2\theta}{2g}=\dfrac{u^2\sin2\theta}{4g}$

u is speed of projection

So,

$\sin^2\theta=\dfrac{2\times \sin\theta \cos\theta}{2}\\\\\sin\theta=\cos\theta\\\\\tan\theta=1\\\\\theta=45^{\circ}$

So, kinetic energy is minimum at maximum height. The velocity at maximum height is, $v\cos\theta$. So, kinetic energy is :

$K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}m(u\cos\theta)^2\\\\K=\dfrac{1}{2}m^2u^2\cos^2\theta\\\\K=\dfrac{1}{2}m^2u^2\cos^2(45)\\\\K=\dfrac{1}{2}m^2u^2\times \dfrac{1}{2}\\\\K=\dfrac{1}{4m}m^2u^2\\\\K=\dfrac{p^2}{4m}$

So, its minimum kinetic energy in its path will be $\dfrac{p^2}{4m}$