1. Use the formula for maximum height of a projectile

For a projectile thrown with speed $u$ at angle $\theta$, the maximum height is

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

The question says both projectiles attain the same height, so their maximum heights are equal.

2. Assign data for the two projectiles

Let the speeds of $P_1$ and $P_2$ be:

$u_1 : u_2 = \sqrt{3} : \sqrt{2}$

So,

$\frac{u_1^2}{u_2^2} = \frac{3}{2}$

Also, for $P_2$,

$\theta_2 = 60^\circ$

Let the angle of projection of $P_1$ be $\theta_1$.

3. Equate maximum heights

Since heights are same,

$\frac{u_1^2 \sin^2\theta_1}{2g} = \frac{u_2^2 \sin^2 60^\circ}{2g}$

Cancel $2g$:

$u_1^2 \sin^2\theta_1 = u_2^2 \sin^2 60^\circ$

Now substitute

$\frac{u_1^2}{u_2^2} = \frac{3}{2}, \qquad \sin^2 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$

So,

$\frac{3}{2} \sin^2\theta_1 = \frac{3}{4}$

Hence,

$\sin^2\theta_1 = \frac{3}{4} \cdot \frac{2}{3} = \frac{1}{2}$

Therefore,

$\sin\theta_1 = \frac{1}{\sqrt{2}}$

So,

$\theta_1 = 45^\circ$

4. Check options

• A: $15^\circ$ ❌
• B: $30^\circ$ ❌
• C: $45^\circ$ ✅
• D: $60^\circ$ ❌

Thus, the correct option is C.