Question

A particle is projected making an angle of ${45}^{\circ }$ with horizontal, with kinetic energy $K$. The kinetic energy at highest point will be

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

Answer 1

The correct option is B $\frac{K}{2}$


Kinetic energy at initial point: $K_1=\frac{1}{2}m{v}^2...\left(1\right)$
Kinetic energy at highest point: $K_2=\frac{1}{2}m{v}^{\prime 2}\text{Velocity at highest point},v^{\prime }=\frac{v}{\sqrt{2}}{K}_2=\frac{1}{2}m[\frac{v}{\sqrt{2}}{]}^2=\frac{1}{4}m{v}^2...(2)$

On comparing eq.(1) and eq. (2),
$K_2=\frac{1}{2}{K}_1$
or $K_2=\frac{1}{2}K$