JEE Physics 2025 Question with Solution

Given: A projectile is launched with speed $u$. Its horizontal range is three times its maximum height.

Find: The value of $n$ in $R = \frac{nu^2}{25g}$.

For projectile motion,

$$R = \frac{u^2 \sin 2\theta}{g}$$

and

$$H = \frac{u^2 \sin^2 \theta}{2g}$$

Given that

$$R = 3H$$

Substituting the formulas,

$$\frac{u^2 \sin 2\theta}{g} = 3\left(\frac{u^2 \sin^2 \theta}{2g}\right)$$

Cancelling $\frac{u^2}{g}$,

$$\sin 2\theta = \frac{3}{2}\sin^2 \theta$$

Using $\sin 2\theta = 2\sin\theta\cos\theta$,

$$2\sin\theta\cos\theta = \frac{3}{2}\sin^2\theta$$

So,

$$4\cos\theta = 3\sin\theta$$

Hence,

$$\tan\theta = \frac{4}{3}$$

Therefore, from the $3\text{-}4\text{-}5$ triangle,

$$\sin\theta = \frac{4}{5}, \qquad \cos\theta = \frac{3}{5}$$

Now,

$$R = \frac{u^2 \sin 2\theta}{g} = \frac{u^2(2\sin\theta\cos\theta)}{g}$$

Substituting the values,

$$R = \frac{u^2\left(2 \times \frac{4}{5} \times \frac{3}{5}\right)}{g} = \frac{24u^2}{25g}$$

Comparing with

$$R = \frac{nu^2}{25g}$$

we get

$$n = 24$$

Therefore, the correct option is D.

Given: $R = 3H$ for a projectile with initial speed $u$.

Find: The value of $n$ if $R = \frac{nu^2}{25g}$.

Start with

$$\frac{R}{H} = 3$$

Using the standard expressions,

$$\frac{\frac{u^2\sin 2\theta}{g}}{\frac{u^2\sin^2\theta}{2g}} = 3$$

This gives

$$\frac{2\sin 2\theta}{\sin^2\theta} = 3$$

Now use $\sin 2\theta = 2\sin\theta\cos\theta$,

$$\frac{2(2\sin\theta\cos\theta)}{\sin^2\theta} = 3$$

So,

$$\frac{4\cos\theta}{\sin\theta} = 3$$

Hence,

$$\cot\theta = \frac{3}{4} \qquad \text{or} \qquad \tan\theta = \frac{4}{3}$$

Thus,

$$\sin\theta = \frac{4}{5}, \qquad \cos\theta = \frac{3}{5}$$

and therefore

$$\sin 2\theta = 2\sin\theta\cos\theta = 2\times \frac{4}{5} \times \frac{3}{5} = \frac{24}{25}$$

So the range is

$$R = \frac{u^2}{g}\cdot \frac{24}{25} = \frac{24u^2}{25g}$$

Comparing with $\frac{nu^2}{25g}$, we obtain $n = 24$.

Therefore, the correct option is D.