Question

A particle is projected from the ground with a kinetic energy E at an angle of ${60}^{\circ }$ with the horizontal. Its kinetic energy at the highest point of its motion will be :

• A. $\frac{E}{\sqrt{2}}$
• B. $\frac{E}{2}$
• C. $\frac{E}{4}$
• D. $\frac{E}{8}$

Answer 1

Answer: (C)

$\frac{E}{4}$

At the highest point vertical component of velocity will be zero, only horizontal component will remain. Let horizontal component $=u$ and speed of projection $=U$.

$u=U\cos 60°=\frac{U}{2}$

Since speed is halved, kinetic energy will become one-fourth. So, kinetic energy at highest point will be $\frac{E}{4}$.