Question

A gun has muzzle speed \({\rm{150m/s}}\). Find two angles of elevation that can be used to hit a target 800 m away.

Short answer

The angles of elevation are \({10^0}or{\rm{ }}{80^0}\).

Step-by-step solution

Range of the projectile

From the trajectory equation the range or the horizontal distance traveled by the projectile is given as-

\(R = \frac{{v_0^2\sin 2\alpha }}{g}\)

where \({v_0}\)is the initial velocity, g is the acceleration of gravity.

Finding speed of the fired gun

The muzzle speed of the gun is \(150m/s\).

Equations of motion can be written as-

\(\begin{aligned}{l}x = 150\cos \alpha t\\y = 150\sin \alpha t - \frac{1}{2}g{t^2}\end{aligned}\)

Vertical acceleration will be constant which is \( - 9.8m/{s^2}\).

Using formula of distance traveled,

\(d = \frac{{v_0^2\sin 2\alpha }}{g}{\rm{ \_\_\_\_}}E{q^n}{\rm{ }}1\)
Substitute \(d = 800,{v_0} = 150,g = 9.8\)

\(E{q^n}{\rm{ }}1\)can be written as-

\(\begin{aligned}{c}800 = \frac{{{{150}^2}\sin 2\alpha }}{{9.8}}\\\sin 2\alpha = \frac{{800 \cdot 9.8}}{{{{150}^2}}}\\ = 0.348\end{aligned}\)

\(\begin{aligned}{c}2\alpha = {\sin ^{ - 1}}(0.348) \vee 2\alpha = {180^^\circ } - {\sin ^{ - 1}}(0.348)\\2\alpha \approx {20^^\circ } \vee 2\alpha \approx {180^^\circ } - {20^^\circ }\\\alpha \approx {10^^\circ } \vee \alpha \approx {80^^\circ }\end{aligned}\)

Therefore, the angles of elevation are \({10^0}or{\rm{ }}{80^0}\).