JEE Physics 2025 Question with Solution

Given: A ball has initial kinetic energy $KE$ and is projected at an angle of $60^\circ$ from the horizontal.

Find: The kinetic energy of the ball at the highest point of its flight.

At the highest point of projectile motion, the vertical component of velocity becomes zero, so only the horizontal component remains.

The initial kinetic energy is

$$KE = \frac{1}{2}mv^2$$

If the initial speed is $v$, then the horizontal component is

$$v_x = v\cos 60^\circ = \frac{v}{2}$$

At the highest point, the speed of the ball is only $\frac{v}{2}$.

Therefore, the kinetic energy at the highest point is

$$KE_{\text{highest}} = \frac{1}{2}m\left(\frac{v}{2}\right)^2$$ $$= \frac{1}{2}m \cdot \frac{v^2}{4}$$ $$= \frac{1}{4}\left(\frac{1}{2}mv^2\right)$$ $$= \frac{KE}{4}$$

Therefore, the kinetic energy at the highest point is $\frac{KE}{4}$. The correct option is B.

Given: Initial kinetic energy is $KE$ and the projection angle is $60^\circ$.

Find: Kinetic energy at the highest point.

Since kinetic energy depends on the square of speed, at the highest point we only need the horizontal component of velocity.

The horizontal speed is

$$v\cos 60^\circ = \frac{v}{2}$$

So the new kinetic energy becomes the initial kinetic energy multiplied by

$$\left(\cos 60^\circ\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$

Hence,

$$KE_{\text{highest}} = KE \cdot \frac{1}{4} = \frac{KE}{4}$$

Therefore, the correct option is B.