A body falling from a high Minaret travels 40 meters in the last 2 seconds of its fall to ground. Height of Minaret in meters is (take g=10 $m / s^{2}$ )

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Solution

The correct option is B 45
Let height of minarate is H nad bodt take T to fall from top to bottom.

$H = \frac{1}{2} g T^{2} . . . (i)$
In last 2 sec,body travels distance of 40 metres so in (T-2)sec distance
travelled =(H-40)m.
$(H - 40) = \frac{1}{2} g (T - 2)^{2} . . . (i i)$
By solving (i) and (ii) T=3 sec and H=45 m.

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A body falling from a high Minaret travels 40 meters in the last 2 seconds of its fall to ground. Height of Minaret in meters is (take g=10 m/s2)

The correct option is B 45Let height of minarate is H nad bodt take T to fall from top to bottom. H=12gT2 ...(i) In last 2 sec,body travels distance of 40 metres so in (T-2)sec distance travelled =(H-40)m. (H−40)=12g(T−2)2 ...(ii) By solving (i) and (ii) T=3 sec and H=45 m.

A body falling from a high Minaret travels 40 meters in the last 2 seconds of its fall to ground. Height of Minaret in meters is (take g=10 $m / s^{2}$ )