A stone is dropped from the top of a tower. When it crosses a point 5 m below the top, another stone is let fall from a point 25 m below the top. Both stones reach the bottom of the tower simultaneously. Find the height of the tower. Take $g = 10 m / s^{2} .$

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Solution

$d i s p l a c e m e n t o f t h e f i r s t s t o n e w h e n s e c o n d i s d r o p p e d = 5 m, t h u s t i m e d i f f e r e n c e i n d r o p p i n g o f t h e s t o n e s i s t = \frac{2 h}{g} = 1 s h = \frac{1}{2} g t^{2} = \frac{1}{2} g {(t - 1)}^{2} + 25 \Rightarrow t = 3 s t h u s, h = 45 m$

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A stone is dropped from the top of a tower. When it crosses a point 5 m below the top, another stone is let fall from a point 25 m below the top. Both stones reach the bottom of the tower simultaneously. Find the height of the tower. Take g=10m/s2.

displacementofthefirststonewhensecondisdropped=5m,thustimedifferenceindroppingofthestonesist=2hg=1sh=12gt2=12g(t−1)2+25⇒t=3sthus,h=45m

A stone is dropped from the top of a tower. When it crosses a point 5 m below the top, another stone is let fall from a point 25 m below the top. Both stones reach the bottom of the tower simultaneously. Find the height of the tower. Take $g = 10 m / s^{2} .$