a stone is dropped from the …
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Payal Singh 8 years, 6 months ago
Let height of tower be 'h'
Time taken for 1st stone to reach ground = {tex}\sqrt{2h\over g}{/tex}
{tex}\left[ s = ut+ {1\over 2}at^2 ;h = {1\over 2}gt^2\right]{/tex}
Time taken for the first stone to be travel 5 m from top = {tex}\sqrt{2\times 5\over g}{/tex}
Time taken for 2nd stone to reach ground from a point (h−25)m above the ground = {tex}\sqrt{2(h-25)\over g}{/tex}
So,
{tex} \sqrt{2h\over g} -\sqrt{2\times 5\over g} = \sqrt{2(h-25)\over g}{/tex}
{tex}=> \sqrt{2h} -\sqrt{10} = \sqrt{2(h-25)}{/tex}
Squaring both sides
=> 2h + 10 - {tex}2\sqrt {20h}= {/tex}2h -50
=> {tex}\sqrt{20h} = 30{/tex}
Sqauring both sides
=> 20h = 900
=> h = 45
0Thank You