Question

A ball whose kinetic energy is E is thrown at an angle of ${45}^{\circ }$ with the horizontal. Its kinetic energy at the highest point of its trajectory will be:

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

Answer 1

The correct option is C

$\frac{E}{2}$

Let m be mass of the ball. If v is its speed of projection, its kinetic energy is
$E=\frac{1}{2}m{v}^2$
At the highest point, the velocity of the particle is $v^{\prime }=vcos\theta =vcos{45}^{\circ }=\frac{v}{\sqrt{2}}.$ Therefore its kinetic energy at this point is
$E^{\prime }=\frac{1}{2}m{v}^{\prime 2}=\frac{1}{2}m{\left(\frac{v}{\sqrt{2}}\right)}^2=\frac{1}{2}\times \frac{1}{2}m{v}^2=\frac{E}{2}$
Hence, the correct choice is (c).