Question

A machine gum is moved on the top of a tower 100 m high. At what angle should the gum the inclined to cover a maximum rages of ** on the ground below ? The muzzle speed of the bullet is 150 ms-1 (Take g = 10 ms-2).

Answer 1

Here, u = 150 ms-1

g = 10 ms-2

Let θ be the angle for maximum range, then the horizontal component of velocity = 150 cosθ and the vertical component of velocity = 150 sinθ.

If T = Time of flight,then

horizontal range = R = (150 cosθ) x T.......(i)

For vertical motion,

Solving the equation,

By using equation (ii) in equation (i), we get

R = 150 cosθ (15 cosθ + √(225 sin2θ +20))

By using various values of θ around 45°, the value of θ comes out to be 43.8° for maximum range.