Question

A machine gun is mounted on the top of a tower 100 m high. At what angles should the gun be inclined to cover a maximum range of firing on the ground below? The muzzle speed of bullet is 150 m/s. Take g= 10 m/s².

Answer 1

Step I : Horizontal range

R = ucos θ × t

where θ is the angle of inclination of gun to cover maximum range and t is the time. i.e., R = 150 cos θ × t   …(i) 

Step II : Vertical height,

H = usin θt + \(\frac{1}{2}gt^2\)

i.e., 100 = −150 sin θt − \(\frac{1}{2}\times10\times t^2\)

t² – (30 sin θ)t – 20 = 0

By using x = \(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) for ax² + bx + c 

we get, t = \(\frac{30sin\,\theta\pm\sqrt{900sin^2\,\theta-1\times20\times1}}{1}\)

= 15sin θ ± \(\sqrt{225\,sin^2\theta-20}\)  …(ii) 

Step III : Putting the value of t from eqn. (ii) in eq, (i), we get,

R = 150 cos θ (15 sin θ + \(\sqrt{225\,sin^2\theta+20}\))

Putting various values of θ around 45º the value of θ comes to be 45.8º for maximum range.

(∵ θ = 45º for maximum range had the shot been fired from ground itself.)