Question

A particle is projected making angle ${45}^o$ with horizontal having kinetic energy $K$. the kinetic energy at highest point will be

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

Answer 1

The correct option is B $K/2$

$K=\frac{1}{2}m{u}^2$

At heightst point velocity will become $\frac{u}{\sqrt{2}}$

$K^{\prime }=\frac{1}{2}m\times (\frac{u}{\sqrt{2}}{)}^2$

$K^{\prime }=\frac{K}{2}$