Step: Find the range of a projectile.
The motion of the projectile is given by;

Given, \(\theta=15^{\circ}\) and \(R=50 ~\text{m}\)
The range of a projectile is given by;
\({R}=\frac{u^2 \sin 2 \theta}{g}\)
Putting all the given values in the formula, we get;
\(\Rightarrow{R}=50~\text{m}=\frac{u^2 \sin \left(2 \times 15^{\circ}\right)}{g} \)
\(\Rightarrow 50 \times g=u^2 \sin 30^{\circ}=u^2 \times \frac{1}{2} \)
\(\Rightarrow 50 \times g \times 2=u^2 \)
\(\Rightarrow u^2=50 \times 9.8 \times 2=100 \times 9.8=980 \)
\(\Rightarrow u=\sqrt{980} \)
\(\Rightarrow u=\sqrt{49 \times 20}\)
\( \Rightarrow u=7 \times 2 \times \sqrt{5}~\text{m/s}\)
\(\Rightarrow u=31.304~\text{m/s}\)
\(\text {For } \theta=45^{\circ}, {R}=\frac{u^2 \sin( 2 \times 45^{\circ})}{g}=\frac{u^2}{g}~~~~~~~~~~~~~~\left(\because \sin 90^{\circ}=1\right)\)
\( \Rightarrow {R}=\frac{(14 \sqrt{5})^2}{g} \)
\(\Rightarrow R=\frac{14 \times 14 \times 5}{9.8} \)
\(\Rightarrow R=100~\text{m}\)
Hence, option (3) is the correct answer.
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