Question

A gun fires a bullet at a speed of $140m{s^{ - 1}}$ . If the bullet is to hit a target at the same level as the gun and at $1km$ distance, the angle of projection may be;
A. ${60^0}$ or ${30^0}$
B. ${40^0}$ or ${50^0}$
C. ${15^0}$ or ${75^0}$
D. ${20^0}$ or ${70^0}$

Answer 1

Hint: For a given horizontal range there are two angles of projections (except maximum range) i.e., $\theta $ or $\left( {{{90}^0} - \theta } \right)$. The horizontal range is given as, $X = \dfrac{{{u^2}\sin 2\theta }}{g} = \dfrac{{{u^2}\sin \left( {{{180}^0} - 2\theta } \right)}}{g}$.
Where, $u$ is the initial velocity, $\theta $ is the angle of projection and $g = 9.8m{s^{ - 1}}$.

Complete step by step solution:
It is given that the initial speed of the bullet is $u = 140m{s^{ - 1}}$.
The horizontal range $X = 1km = {10^3}m$.
We know that the horizontal range, $X = \dfrac{{{u^2}\sin 2\theta }}{g}$.
Substitute all the required values in the above formula.
\[ \Rightarrow {10^3} = \dfrac{{{{\left( {140} \right)}^2}\sin 2\theta }}{{9.8}}\]
Further simplifying
$ \Rightarrow \sin 2\theta = \dfrac{1}{2}$
$ \Rightarrow 2\theta = {30^0}$
$\theta = {15^0}$
Another possible angle of projection is $\left( {90 - \theta } \right) = 90 - {15^0} = {75^0}$.
Hence, the correct option is (c) ${15^0}$ or ${75^0}$.

Note:
A body projected into space and is no longer being propelled by fuel is called a projectile. The horizontal range $\left( X \right)$ of a projectile depends upon initial speed $\left( u \right)$ and angle of projection $\left( \theta \right)$. For $\theta = {45^0}$, the horizontal range is maximum.