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  4. An aeroplane flying horizontally with a speed of 360 km 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

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Q. An aeroplane flying horizontally with a speed of $360\,km\,h^{-1}$ 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

Motion in a Plane

Solution:

Time taken by the bomb to fall through a height of $490\,m$
$t=\sqrt{\frac{2h}{g}}$
$=\sqrt{\frac{2\times490}{9.8}}=10\,s$
Distance at which the bomb strikes the ground
= horizontal velocity $\times$ time
$=360\,km\,h^{-1}\times10\,s$
$=360\,km\,h^{-1}\times \frac{10}{3600}h$
$=1\,km$