Question

A particle is projected at an angle $\theta$ from the horizontal with kinetic energy $K$. What is the kinetic energy of the particle at the highest point?

Answer 1

Given, Kinetic energy $K$, Let velocity of particle is $v$.

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

Velocity in horizontal direction$=vcos\left(\theta \right)=\sqrt{\frac{2K}{m}}cos\left(\theta \right)$

Velocity in vertical direction$=vsin\left(\theta \right)=\sqrt{\frac{2K}{m}}sin\left(\theta \right)$

At highest point, Vertical velocity is zero only constant horizontal velocity is present.

Kinetic energy at highest point $=\frac{1}{2}m\left(vcos\right(\theta ){)}^2=\frac{1}{2}\times m\times \frac{2K}{m}co{s}^2(\theta )=Kco{s}^2(\theta )$