JEE Physics 2023 Question with Solution

Given: Angle of projection in the first case is $\theta_1 = 15^\circ$, range is $R_1 = 50 \, \text{m}$, and in the second case $\theta_2 = 45^\circ$. The initial velocity $u$ remains the same.

Find: The range at $45^\circ$.

For projectile motion, the range is

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

So, for the two cases,

$$\frac{R_2}{R_1} = \frac{\sin(2\theta_2)}{\sin(2\theta_1)}$$

Substituting $\theta_1 = 15^\circ$ and $\theta_2 = 45^\circ$,

$$\frac{R_2}{50} = \frac{\sin(90^\circ)}{\sin(30^\circ)}$$

Using $\sin 90^\circ = 1$ and $\sin 30^\circ = \frac{1}{2}$,

$$\frac{R_2}{50} = \frac{1}{1/2} = 2$$

Hence,

$$R_2 = 50 \times 2 = 100 \, \text{m}$$

Therefore, the range of the projectile is $100 \, \text{m}$. The correct option is C.

Given: In the first case, $R = 50 \, \text{m}$ at $15^\circ$. In the second case, the projectile is projected at $45^\circ$ with the same velocity.

Find: The new range.

Using the range formula,

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

For $\theta = 15^\circ$,

$$50 = \frac{u^2 \sin 30^\circ}{g}$$ $$50 = \frac{u^2}{2g}$$

So,

$$\frac{u^2}{g} = 100$$

Now for $\theta = 45^\circ$,

$$R = \frac{u^2 \sin 90^\circ}{g}$$ $$R = \frac{u^2}{g}$$

Using $\frac{u^2}{g} = 100$,

$$R = 100 \, \text{m}$$

Therefore, the range at $45^\circ$ is $100 \, \text{m}$. The listed the solution shows "The Correct Option is B", but the worked calculation gives $100 \, \text{m}$, which matches option C.