- Home
- Standard 11
- Physics
3-2.Motion in Plane
medium
A fighter plane flying horizontally at an altitude of $1.5\; km$ with speed $720\; km / h$ passes directly overhead an anti-atrcraft 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? At what minimum altitude should the pilot fly the plane to avoid being hit ? (Take $g=10 \;m s ^{-2}$ ).
A$38.5^o\;and \,12\;km$
B$29.5^o\;and \,16\;km$
C$19.5^o\;and \,20\;km$
D$19.5^o\;and\, 16\;km$
Solution
Height of the fighter plane $=1.5 \,km =1500 \,m$Speed of the fighter plane, $v=720 \,km / h =200 \,m / s$
Let $\theta$ be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.
Muzzle velocity of the gun, $u=600 \,m / s$ Time taken by the shell to hit the plane $=t$ Horizontal distance travelled by the shell $=u_{x} t$ Distance travelled by the plane $=v t$ The shell hits the plane. Hence, these two distances must be equal.
$u_{ x } t=v t$
$u \sin \theta=v$
$\sin \theta=\frac{v}{u}$
$=\frac{200}{600}=\frac{1}{3}=0.33$
$\theta=\sin ^{-1}(0.33)$
$=19.5^o$
In order to avoid being hit by the shell, the pilot must fly the plane at an altitude $(H)$ higher than the maximum height achieved by the shell.
$\therefore H=\frac{u^{2} \sin ^{2}(90-\theta)}{2 g }$
$=\frac{(600)^{2} \cos ^{2} \theta}{2 g }$
$=\frac{360000 \times \cos ^{2} 19.5}{2 \times 10}$
$=18000 \times(0.943)^{2}$
$=16006.482 \,m$
$\approx 16\; km$
Standard 11
Physics
