Question

A cricket ball of mass m is hitted at the angle ${45}^o$ to the horizontal with velocity v.Its kinetic energy at the topmost point is

• A. $0$
• B. $\frac{1}{2}m{v}^2$
• C. $\frac{m{v}^2}{4}$
• D. $\frac{m{v}^2}{2\sqrt{2}}$

Answer 1

The correct option is C $\frac{m{v}^2}{4}$

At topmost point it wqill have only horizontal velocity which is $V_x$

$K.E$ $=\frac{1}{2}m{v}^2$

$=\frac{1}{2}m\times v^2{\cos }^2\theta$

$=\frac{1}{2}m\times v^2\times {\cos }^2{45}^0=\frac{m{v}^2}{4}.$

Hence,

option $\left(C\right)$ is correct answer.