At the height 80 m, an aeroplane is moving with $150\ {ms}^{- 1}$. A bomb is dropped from it, so as to hit a target. At what distance from the target should bomb be dropped? $(g = 10\ {ms}^{- 2})$

Question

Dr. Abubacker · Accepted Answer

(b) Time taken by the bomb to reach the ground is given by
$h_{OA} = \frac{1}{2}{gt}_{OB}^{2}$
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$we\ have\ \ \ \ \ t_{OB} = \sqrt{\frac{{2h}_{OA}}{g}}$
$given,\ \ \ h_{OA} = 80m,\ \ g = {10\ ms}^{- 2}$
$	herefore\ \ \ \ t_{OB} = \sqrt{\frac{2 	imes 80}{10}\ 4s}$
$Horizontal\ velocity\ of\ bomb\ $
$v = 150\ {ms}^{- 1}$
Horizontal distance covered by the bomb
$AB = {vt}_{OB}$
$= 150 	imes 4$
$= 600\ m$
Hence, the bomb should be dropped 600 m before the target.

Q. At the height 80 m, an aeroplane is moving with $150\ {ms}^{- 1}$. A bomb is dropped from it, so as to hit a target. At what distance from the target should bomb be dropped? $(g = 10\ {ms}^{- 2})$

Explanation

Reviewed by Dr. Abubacker

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