Question

A gun is held at an angle of ${45}^{\circ }$ above the horizontal.
(i) If the range of the projectile is $25m$, what is its muzzle velocity?
(ii) At what angle should the gun be held so that the range is reduced to $18m$?

Answer 1

given holding angle $\theta ={45}^{\circ }$

(1) Range $R=25m$

as we know that $R=\frac{u^2\sin 2\theta }{g}$

$\Rightarrow 25=\frac{u^2\sin {90}^{\circ }}{10}\Rightarrow u=5\sqrt{10}m$

(2) now given that range is reduced to $18m$

from the formula of range

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

$\Rightarrow 18=\frac{{\left(5\sqrt{10}\right)}^2\sin 2\theta }{10}\Rightarrow \theta \simeq {23}^{\circ }$