The range of the projectile projected at an angle of 15 ∘ with horizontal is 50   m . If the…

The data given is 

$$\begin{gathered} {\theta }_1=15° \\ {\theta }_2=45° \\ {R}_1=50 \mathrm{m} \end{gathered}$$

The formula for range of a projectile is given by 

$R=\frac{u^2 \sin 2\theta }{g} ...\left(i\right)$

Substituting the values in equation (i)

$$\begin{gathered} R_1=\frac{u^2\sin 2{\theta }_1}{g} \\ \Rightarrow 50=\frac{u^2\sin 30°}{g} \\ \Rightarrow \frac{u^2}{g}=100 \end{gathered}$$

Let the new range be $R_2$. The magnitude of the range is 

$$\begin{gathered} R_2=\frac{u^2\sin 2{\theta }_2}{g} \\ \Rightarrow R_2=\left(\frac{u^2}{g}\right)\sin 90° \\ \Rightarrow R_2=100 \mathrm{m} \end{gathered}$$