A machine gun is mounted on the top of a tower `60 m` At what angle showld the gun be inclined to cver a vaximu range of firing on the groun below? The muzzle speed of the bullet is `90 ms^(-1)` take ` g= 10 ms^(1)`.

Explanation

Let the machine gun be mounted at (O) and fired with a speed of `90 ms^(-1)` at an angle `theta` with the horizontal. Fig. 2 (d) . 27 .
Total range of firing on ground.
`R' -R+ x =(u^(2) sin 2 theta)/g + 60 cat theta`
.
`(dR')/(theta) =(u ^(2) 2 cos 2 theta)/g + 60 xx (- casec^(2) theta)`
``=(2 u^(2)/g (1-2 sin^(2) theta) -(60)/(sing^(2) theta)`
`=(2 xx 90^(2))/(10) (-2 sin^(2) theta) (60)/(sin^(2) theta)`
`=1620 -3240 sin ^(2) theta -(60) /(sin^(2) theta)_`
Fpr `R' to be maximum, ` (dR') (d theta) =0`
:. `16 20 -3240 som^(2) theta + 1 =0`
or `54 sin^(2) theta + 1=0` ltvegt or ` sin^(2) theta =(27 +- sqrt(-27)^(2) -4 xx 54xx 1))/(2 xx 54)`
`=(27 =-(22.65)/(108)`
Taking only positive sing.
`sing ^(2) theta (49 .65)/(108) =0 . 4597`
`sin `theta =0.6780 or theta =42^(@) 41'`.