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Byju's Answer

Standard XIII

Physics

2nd Equation of Motion

An aeroplane ...

Question

An aeroplane is travelling horizontally at a height of $2000 m$ from the ground. The aeroplane, when at a point P, drops a bomb to hit a stationary target Q on the ground. In order that the bomb hits the target, what angle $θ$ must the line PQ make with the vertical ? $[g = 10 m s^{- 2}]$

15^{\circ}

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30^{\circ}

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90^{\circ}

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45^{\circ}

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Solution

The correct option is D $45^{\circ}$
Let $t$ be the time taken by bomb to hit the target.
$h = 2000 = \frac{1}{2} g t^{2} \Rightarrow t = 20$ sec
$R = u t = (100) (20) = 2000$ m
$∵ tan θ = \frac{R}{h} = \frac{2000}{2000} = 1 \Rightarrow θ = 45^{\circ}$

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Similar questions

Q. An aeroplane is travelling horizontally at a height of

2000 m

from the ground. The aeroplane, when at a point

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, drops a bomb to hit a stationary target

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on the ground. In order that the bomb hits the target, what angle

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[g = 10 {m/s}^{2}]

Q. An aeroplane is travelling horizontally at a height of

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θ

must the line PQ make with the vertical ?

[g = 10 m s^{- 2}]

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At the height 80 m, an aeroplane is moving with 150 m/s. A bomb is dropped from it so as to hit a target. At what distance from the target should the bomb be dropped (given g = 10 $m / s^{2}$ )

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An aeroplane is travelling horizontally at a height of 2000 m from the ground. The aeroplane, when at a point P, drops a bomb to hit a stationary target Q on the ground. In order that the bomb hits the target, what angle θ must the line PQ make with the vertical ? [g=10ms−2]

The correct option is D 45∘Let t be the time taken by bomb to hit the target. h=2000=12gt2⇒t=20 sec R=ut=(100)(20)=2000 m ∵tanθ=Rh=20002000=1⇒θ=45∘

An aeroplane is travelling horizontally at a height of $2000 m$ from the ground. The aeroplane, when at a point P, drops a bomb to hit a stationary target Q on the ground. In order that the bomb hits the target, what angle $θ$ must the line PQ make with the vertical ? $[g = 10 m s^{- 2}]$

The correct option is D $45^{\circ}$
Let $t$ be the time taken by bomb to hit the target.
$h = 2000 = \frac{1}{2} g t^{2} \Rightarrow t = 20$ sec
$R = u t = (100) (20) = 2000$ m
$∵ tan θ = \frac{R}{h} = \frac{2000}{2000} = 1 \Rightarrow θ = 45^{\circ}$