Question

A cricket ball is hit for a six leaving the bat at an angle of ${45}^o$ to the horizontal with kinetic energy $K$. At the top, Kinetic energy of the ball is:

• A. $0$
• B. $K$
• C. $\frac{K}{2}$
• D. $\frac{K}{\sqrt{2}}$

Answer 1

Answer: (C)

$\frac{K}{2}$

Option C is correct

At the top position, the ball contains only horizontal

Velocity, hence $v_{top}=vcos{45}^o$

Initial velocity $=v$.

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

Final $KE=\frac{1}{2}m(vcos{45}^0{)}^2=\frac{\frac{1}{2}m{v}^2}{2}=\frac{K}{2}$