A stone is dropped from the top of the height $h$ . $^{'} n^{'}$ second letter, a second stone is projected downward from the same cliff with a vertically downward velocity $u .$ Show that the two stones will reach the bottom of the cliff together, if $8 h {(u - g n)}^{2} = g n^{2} {(2 u - g n)}^{2}$
What can you say about the limiting value of $^{'} n^{'}$ ?

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Solution

$- h = - \frac{1}{2} g t_{1} 2$
$- h = - 4 t_{2} - \frac{1}{2} g t_{2}$
$t_{1} = \sqrt{\frac{2 h}{g}}$
$t_{2} = t_{1} - n$
$h = (\sqrt{\frac{2 h}{g}} - n) + \frac{1}{2} g (\sqrt{\frac{2 h}{g}} - n)$
$\Rightarrow h = 4 \sqrt{\frac{2 h}{g}} - 4 n + h + \frac{n^{2} g}{2} - g n 4 \sqrt{\frac{2 h}{g}}$
$\Rightarrow 4 \sqrt{\frac{2 h}{g}} (g n - u) = \frac{n}{2} (- 2 u + g)$
$\Rightarrow 8 h {(g n - u)}^{2} - n^{2} g {(2 u - g)}^{2}$

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−h=−12gt12−h=−4t2−12gt2t1=√2hgt2=t1−nh=(√2hg−n)+12g(√2hg−n)⇒h=4√2hg−4n+h+n2g2−gn4√2hg⇒4√2hg(gn−u)=n2(−2u+g)⇒8h(gn−u)2−n2g(2u−g)2