QJEEM25Ph1ace: JEE Main 2025 Physics - Class 11 JEEM25Ph1ace Solution: A ball having kinetic energy KE, is projected at an angle of…

Let's break down the problem step by step:

The ball is projected with kinetic energy

$\mathrm{KE} = \frac{1}{2} m v^2,$

where $v$ is the initial speed.

The ball is launched at an angle of $60^\circ$ . So, its initial velocity components are:

Horizontal: $v_x = v \cos(60^\circ) = \frac{v}{2}$

Vertical: $v_y = v \sin(60^\circ)$

At the highest point of its flight, the vertical component of the velocity becomes zero (i.e., $v_y = 0$ ), but the horizontal component remains unchanged.

Therefore, the kinetic energy at the highest point is only due to the horizontal velocity:

$\mathrm{KE}_{\text{top}} = \frac{1}{2} m v_x^2$

Substitute $v_x = \frac{v}{2}$ :

$\mathrm{KE}_{\text{top}} = \frac{1}{2} m \left(\frac{v}{2}\right)^2 = \frac{1}{2} m \frac{v^2}{4} = \frac{1}{8} m v^2$

Notice that the initial kinetic energy is

$\mathrm{KE} = \frac{1}{2} m v^2.$

So we can write:

$\mathrm{KE}_{\text{top}} = \frac{1}{8} m v^2 = \frac{1}{4} \left(\frac{1}{2} m v^2\right) = \frac{1}{4} \mathrm{KE}$

Thus, the kinetic energy at the highest point is $\frac{\mathrm{KE}}{4}$ .

Answer: Option C.