Question

A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 $km{h}^{-1}$ passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 $m{s}^{-1}$ to hit the plane. (Take g= 10 $m{s}^{-2}$)

• A. sin${}^{-1}$(1/3)
• B. sin${}^{-1}$(2/3)
• C. cos${}^{-1}$(1/3)
• D. cos${}^{-1}$(2/3)

Answer 1

The correct option is A sin${}^{-1}$(1/3)

Here,

Speed of the plane, $v=720km{h}^{-1}=720\times \frac{5}{18}m{s}^{-1}=200m{s}^{-1}$

Speed of the shell, $u=600m{s}^{-1}$Let the shell will hit the plane at L after time t if fired at an angle $\theta$ with the vertical from O.

For hitting, horizontal distance travelled by the plane = horizontal distance travelled by the shell.i.e,

$v\times t=usin\theta \times t$

sin $\theta =\frac{v}{u}$ = $\frac{200m{s}^{-1}}{600m{s}^{-1}}$ = $\frac{1}{3}$

$\theta =si{n}^{-1}\left(\frac{1}{3}\right)$