Question

A body is projected at an angle $\theta$ with the vertical with kinetic energy $K$. What is the kinetic energy of the particle at the highest point?

• A. $K{\cos }^2\theta$
• B. $K{\sin }^2\theta$
• C. $\frac{K}{2}$
• D. $K{\tan }^2\theta$

Answer 1

Answer: (B)

$K{\sin }^2\theta$

At the point of projection,
velocity of the particle $=u$
Thus, $K.E.=K=\frac{m{u}^2}{2}$

Now, At the highest point,
velocity of the particle is is x direction $=u\sin \theta$
Therefore,
$K.E.=\frac{m{u}^2{\sin }^2\theta }{2}=K{\sin }^2\theta$
i.e. option 'B'