Question

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

• A. $\sin ^ { - 1 } \left( \frac { 1 } { 3 } \right)$
• B. $\sin ^ { - 1 } \left( \frac { 2 } { 3 } \right)$
• C. $\cos ^ { - 1 } \left( \frac { 1 } { 3 } \right)$
• D. $\cos ^ { - 1 } \left( \frac { 2 } { 3 } \right)$

Answer 1

Answer: (A)

$\sin ^ { - 1 } \left( \frac { 1 } { 3 } \right)$

Answer 2

Here,

Speed of the plane,

$\mathrm { v } = 720 \mathrm {~km} \mathrm {~h} ^ { - 1 } = 720 \times \frac { 5 } { 18 } \mathrm {~ms} ^ { - 1 }$

$= 200 \mathrm {~ms} ^ { - 1 }$

Speed of the shell,

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.

$\sin \theta = \frac { \mathrm { v } } { \mathrm { u } } = \frac { 200 \mathrm {~ms} ^ { - 1 } } { 600 \mathrm {~ms} ^ { - 1 } } = \frac { 1 } { 3 }$

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