Question

A projectile is thrown at an angle of ${60}^{\circ }$ with horizontal. If $E$ is the kinetic energy with which the projectile is thrown, then find the kinetic energy at the top of trajectory.

• A. $\frac{3E}{4}$
• B. $\frac{E}{4}$
• C. $\frac{E}{2}$
• D. $\frac{\sqrt{3}E}{2}$

Answer 1

The correct option is B $\frac{E}{4}$

Let $u$ be the velocity with which the projectile is thrown.
We know that kinetic energy is given as
$E=\frac{1}{2}m{u}^2$
When the projectile is at the top of its trajectory then its velocity is minimum. Only horizontal component of velocity exists at the top of its trajectory .
$\Rightarrow v_{top}=u\cos \theta$ (where $\theta$ is the angle of projection with the horizontal)
Kinetic energy at the top of trajectory $E_t=\frac{1}{2}m(u\cos \theta {)}^2$
$\Rightarrow E_t=\frac{1}{2}m(u\cos {60}^{\circ }{)}^2$
$\therefore E_t=\frac{E}{4}$