A machine gun is mounted on the top of a tower of height `h`. At what angle should the gun be inclined to cover a maximum range of firing on the ground below ? The muzzle speed of bullet is `150 ms^-1`. Take `g = 10 ms^-2`.

Answer

`43.7`

Explanation

Let u be the muzzle speed of the bullet fired from the gun (on the top of the tower ) at an angle ` theta ` with the horizontal as shown in Figure
Clearly, the total range of firing on the ground is
` x = ( u^(2) sin 2 theta)/( g) + 100 cot theta `
` :. ( dx)/( d theta ) = ( u^(2) xx 2 cos 2 theta )/( g) + 100 xx ( - cosec ^(2) theta)`
`= (2 u^(2))/( g) (1 + 2 sin ^(2) theta) - (100)/( sin ^(2) theta)`
` = 4500 - 9000 sin^(2) theta - (100)/( sin ^(2) theta )`

For x to be maximum
`( dx)/( d theta) = 0`
or `4500 - 9000 sin^(2) theta - (100)/( sin ^(2) theta ) = 0`
or ` 90 sin^(2) theta - 45 sin^(2) theta + 1 = 0`
` sin ^(2) theta = ( 45 pm sqrt((- 45)^(2) - 4 xx 90 xx 1))/(2 xx 90)`
`= ( 45 pm 40.80)/( 180)`
Taking only positive sign,
` sin^(2) theta = 0.4767`
or ` sin theta = 0.6904 or theta = 43 . 7^(@)`