A fightre plane flying horizontally at an altitude oa ` 1.5 km` with speed ` 720 h^(-1)` passes directly over head an anticraft gun. At what angle from the vertical should the gun be fired from the shell with muzzle speed ` 600 ms^(-1)` to hit plane. At what minimum altitude showld the pilot fly the plane to avoid being hit ? ( Take g= 10 ms^(-2)).

Explanation

In Fig. (NCT). 24, (O)
be the position of gun and (A) be the position of plane,
.
The speed of the plance,
` v= ( 720 xx 1000)/( 60 xx 60) = 200 ms^(-1)`
The speed of the shell, ` u = 600 ms^(-1)`
Let the shell will hit the plance at (B) after time (T) if fired at and angle ` theta` with the vertical from (O). Then the horizontal distance travelled by shell in time (t) is the same as the distance covered by the plane ltBrgt i.e., ` u_x xx t vt ` or ` u sin theta t = vt`
or ` sin theta = v/ u = ( 200)/( 600) = 1/ 3 = 0. 333 = sin 1 9 . 5 ^@ ` or ` theta = 19.5^2` with the vertical
The maximum height attained by bullet is,
` H= u^2 sin ^2 ( 90^@ - theta) )/( 2 g) = )u^2 cos^2 theta)/( 2 g)` [ :. sin theta = 1//3 , cos theta = sqrt 8//3 ]`
` = ( ( 600)^@ xx ( sqrt 8 / 30^2)/( 2 xx 10) = 1600 m = 16 km `
The plance will not be hit by the bullet from the gun if it is flying at a minimum height which is more than the maximum height (H) attatined by bullet after firing from gun (i.e, gt 16 km )`.