The range of the projectile projected at an angle of 15 ^ ∘ with...

The data given is

${\theta}_{1} = 15 ° {\theta}_{2} = 45 ° {R}_{1} = 50m$

The formula for range of a projectile is given by

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

Substituting the values in equation (i)

${R}_{1} = \dfrac{{u}^{2}\sin 2{\theta}_{1}}{g} ⇒ 50 = \dfrac{{u}^{2}\sin 30 °}{g} ⇒ \dfrac{{u}^{2}}{g} = 100$

Let the new range be ${R}_{2}$. The magnitude of the range is

${R}_{2} = \dfrac{{u}^{2}\sin 2{\theta}_{2}}{g} ⇒ {R}_{2} = \left(\dfrac{{u}^{2}}{g}\right)\sin 90 ° ⇒ {R}_{2} = 100m$