JEE Physics 2026 Question with Solution

Given: The object is projected from point A with initial kinetic energy $K$ at angle $60^\circ$. Point B is the highest point of the projectile and point C is the landing point.

Find: The ratio of the difference in kinetic energies at points B and C to that at point A.

In projectile motion, the horizontal component of velocity remains constant, while the vertical component becomes zero at the highest point.

At point A,

$$K_A = \frac{1}{2} m v_0^2 = K$$

The horizontal velocity is

$$v_x = v_0 \cos 60^\circ = \frac{v_0}{2}$$

At the highest point B, the vertical component is zero, so kinetic energy is only due to the horizontal component:

$$K_B = \frac{1}{2} m \left(\frac{v_0}{2}\right)^2 = \frac{K}{4}$$

At point C, the projectile returns to the same horizontal level as A, so its speed is again $v_0$ and hence

$$K_C = K$$

Therefore, the difference in kinetic energies at points B and C is

$$K_C - K_B = K - \frac{K}{4} = \frac{3K}{4}$$

Now compare this with the kinetic energy at point A:

$$\frac{\frac{3K}{4}}{K} = \frac{3}{4}$$

Therefore, the required ratio is $3 : 4$. The correct option is C.

Given: Initial kinetic energy is $K$ and launch angle is $60^\circ$.

Find: Ratio of the loss in kinetic energy from the initial point to the highest point, compared with the initial kinetic energy.

At maximum height of a projectile, the kinetic energy is not zero. It becomes

$$K_{\text{top}} = K \cos^2 \theta$$

Here, $\theta = 60^\circ$, so

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

Initial kinetic energy at A is

$$K_A = K$$

Hence, the difference is

$$K_A - K_B = K - \frac{K}{4} = \frac{3K}{4}$$

Now ratio with kinetic energy at A:

$$\frac{\frac{3K}{4}}{K} = \frac{3}{4}$$

So the ratio is $3 : 4$.

The solution states points B and C, but the working clearly evaluates the kinetic energy change between the initial point and the highest point. That gives option C.