First slide

Projection Under uniform Acceleration

Question

A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 $km h^{- l}$ 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 $s^{- l}$ to hit the plane. (Take g: l0 m $s^{- 2}$ )

Moderate

A

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

$\sin^{- 1} (\frac{2}{3})$
C

$\cos^{- 1} (\frac{1}{3})$
D

$\cos^{- 1} (\frac{2}{3})$

Solution

Here,
Speed of the plane, v = 720 km $h^{- 1}$
= $720 \times \frac{5}{18} {ms}^{- 1} = 200 {ms}^{- 1}$
Speed of the shell, u = 600 ${ms}^{- 1}$
Let the shell will hit the plane at L after time t if fired at an angle $θ$ with the vertical from O.
For hitting, horizontal distance travelled by the plane = horizontal distance travelled by the shell.
i.e., $v \times t = u \sin θ \times t$
$sin θ = \frac{v}{u} = \frac{200 {ms}^{- 1}}{600 {ms}^{- 1}} = \frac{1}{3}$
$θ = \sin^{- 1} (\frac{1}{3})$

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