Explanation:
Given : \({\theta_{1}=15^{\circ}, R_{1}=50 {~m}, \theta_{2}=45^{\circ}, R=}\) ? Range of a projectile is\(R = \frac{{{u^2}\sin 2\theta }}{g}\)
\({\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 50 = \frac{{{u^2}\sin 30^\circ }}{g}\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
\({\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {R_2} = {\text{ }}\frac{{{u^2}\sin {\text{ }}90^\circ }}{g}\,\,\,\,\,\,\,\,\,\,\,\,(2)\)
Dividing eq. (1) by eq. (2), we get \({\mkern 1mu} {\mkern 1mu} \,\frac{{50}}{{{R_2}}} = \frac{1}{2} \Rightarrow {R_2} = 100\;m\).
So correct option is (2)