An aeroplane flying horizontally with a speed of 360 $k m h^{- 1}$ releases a bomb at a height of 490 m from the ground. If g = 9.8 $m s^{- 2}$ , it will strike the ground at

10 km

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100 km

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1 km

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16 km

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Solution

The correct option is B 1 km
First convert speed $360 k m / h i n \times m / s$

$360 \times \frac{1000}{3600} = 100 m / s$

When the bomb is dropped, it will have an initial horizontal velocity which is equal to the speed of the plane. So the bomb fall and will travel forward.

Initial horizontal velocity $(v) = 100 m / s$

Vertical velocity $(u) = 0 m / s$

Height from which bomb is dropped $(h) = 490 m$

Time= $t$ secs

Now from 2nd equation of motion $h = u t + \frac{1}{2} g t^{2}$

$490 = 0 (t) + (\frac{1}{2}) \times 9.8 \times t^{2}$

$490 = 4.9 \times t^{2}$

$t^{2} = 100$

$t = 10 s e c s$

So 10secs is the total time which bomb takes to reach the ground by traveling $y$ (let) distance.

So $y = v t = 100 m / s \times 10 s e c s = 1000 m$

Distance travel by bomb $= 1000 m$ or $1 k m$

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An aeroplane flying horizontally with a speed of 360 kmh−1 releases a bomb at a height of 490 m from the ground. If g = 9.8 ms−2, it will strike the ground at

An aeroplane flying horizontally with a speed of 360 $k m h^{- 1}$ releases a bomb at a height of 490 m from the ground. If g = 9.8 $m s^{- 2}$ , it will strike the ground at