Question

The kinetic energy of a projectile at the highest point is half of the initial kinetic energy. What is the angle of projection with the horizontal?

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is B ${45}^{\circ }$

Let $(KE{)}_i$ and $(KE{)}_h$ be the initial kinetic energy and kinetic energy at highest point respectively.
Let $v$ be the initial velocity of projectile.
At highest point, the velocity of projectile $=v\cos \theta$
According to question
$(KE{)}_h=\frac{1}{2}(KE{)}_i$
$\Rightarrow \frac{1}{2}m{v}^2{\cos }^2\theta =\frac{1}{2}\left(\frac{1}{2}m{v}^2\right)$
$\Rightarrow {\cos }^2\theta =\frac{1}{2}$
$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}$
$\Rightarrow \theta ={45}^{\circ }$