1. Use the formula for range of a projectile

For a projectile projected with speed $u$ at angle $\theta$, the horizontal range is

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

2. Given first projection

The projectile is projected at $15^\circ$ and its range is $50\,\text{m}$.

So,

$50 = \frac{u^2 \sin(2\times 15^\circ)}{g}$

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

Since,

$\sin 30^\circ = \frac{1}{2}$

we get

$50 = \frac{u^2}{g}\cdot \frac{1}{2}$

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

3. Now find the range for angle $45^\circ$

Again using

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

For $\theta = 45^\circ$,

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

and

$\sin 90^\circ = 1$

Thus,

$R = \frac{u^2}{g} = 100\,\text{m}$

4. Check options

• A: $50\sqrt{2}$ m
• B: $100$ m
• C: $100\sqrt{2}$ m
• D: $50$ m

Hence, the correct option is B.