Question

A gun is fired with angle of elevation\({\rm{3}}{{\rm{0}}^{\rm{0}}}\). What is the muzzle speed if the maximum height of the shell is 500 m?

Short answer

The muzzle speed of the gun is \(198m/s\).

Step-by-step solution

Third equation of motion

The third equation of motion is \({{\rm{v}}^{\rm{2}}}{\rm{ = }}{{\rm{u}}^{\rm{2}}}{\rm{ + 2as}}\); where u and v are the initial and final velocity. a is the acceleration and s is the distance traveled.

Finding speed of the fired gun

Let the muzzle speed be u.

Initial upward speed can be written as-

\(\begin{aligned}{c}u\sin \theta = u\sin {30^0}\\ = \frac{u}{2}\end{aligned}\)

i.e., the initial vertical velocity is \(\frac{u}{2}\) and final vertical velocity will be 0.

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

Using third equation of motion,

\(\begin{aligned}{c}{v^2} - {u^2} = 2as\\0 - \frac{{{u^2}}}{4} = 2( - 9.8)(500)\\{u^2} = 39200\\u \approx 198\end{aligned}\)

Therefore, the muzzle speed of the gun is \(198m/s\).