If a ball fallen freely from 'h' height reaches in time 't' at ground, then what will be the time when it reaches at height $\frac{h}{2}$ ?

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Solution

At instant initial velocity=0m/s(at the start of fall)
From equation of motion we get,
$v^{2}$ = $u^{2} + 2 g s$
where s $= \frac{h}{2}$
$v$ = $\sqrt{10 h}$
Also
$v$ =ut+ $\frac{1}{2} g t^{2}$
$t^{2}$ = $\frac{2 \sqrt{10 h}}{10}$ = $\frac{2 \sqrt{h}}{\sqrt{10}}$ t= $\sqrt{\frac{2 \sqrt{h}}{\sqrt{10}}} s$

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If a ball fallen freely from 'h' height reaches in time 't' at ground, then what will be the time when it reaches at height h2?

At instant initial velocity=0m/s(at the start of fall)From equation of motion we get,v2=u2+2gswhere s=h2v=√10hAlso v=ut+12gt2t2=2√10h10 =2√h√10t=√2√h√10s

If a ball fallen freely from 'h' height reaches in time 't' at ground, then what will be the time when it reaches at height $\frac{h}{2}$ ?