Question

A ball whose kinetic energy is E, is thrown at an angle of ${45}^o$ with the horizontal, its kinetic energy at the highest point of its flight will be?

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

Answer 1

The kinetic energy of the ball at the highest point is as follows:

the expression for kinetic energy is $KE=\frac{1}{2}m{v}^2$.

So, the kinetic energy at the highest point is,

$K{E}_h=\frac{1}{2}m\left(v\cos \right({45}^{\circ }){)}^2$

$K{E}_h=\frac{1}{2}m\left(v^2\right)\left(\cos \right({45}^{\circ }){)}^2$

$K{E}_h=E\left(\cos \right({45}^{\circ }){)}^2$

$K{E}_h=\frac{E}{2}$

so, at the highest point the kinetic energy is $\frac{E}{2}$.