A stone is dropped from the top of a building | vedclass

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(B) Let the total height of the building be $H$. The first stone is dropped from the top $(u=0)$.When it has fallen $5 \, m$,its velocity $v$ is given

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) Let the total height of the building be $H$. The first stone is dropped from the top $(u=0)$.When it has fallen $5 \, m$,its velocity $v$ is given by $v^2 = u^2 + 2gh = 0 + 2 	imes 10 	imes 5 = 100$,so $v = 10 \, m/s$.The time taken by the first stone to fall $5 \, m$ is $t_1 = v/g = 10/10 = 1 \, s$.At this instant,the second stone is dropped from a point $25 \, m$ below the top. Let the total time taken by the first stone to reach the ground be $T$. Then the time taken by the second stone is $(T - 1) \, s$.The distance covered by the first stone is $H = \frac{1}{2} g T^2$.The distance covered by the second stone is $H - 25 = \frac{1}{2} g (T - 1)^2$.Subtracting the two equations: $H - (H - 25) = \frac{1}{2} g [T^2 - (T - 1)^2] \Rightarrow 25 = 5 [T^2 - (T^2 - 2T + 1)] \Rightarrow 25 = 5(2T - 1) \Rightarrow 5 = 2T - 1 \Rightarrow 2T = 6 \Rightarrow T = 3 \, s$.Substituting $T = 3 \, s$ into the first equation: $H = \frac{1}{2} 	imes 10 	imes (3)^2 = 5 	imes 9 = 45 \, m$.

$A$ stone is dropped from the top of a building. When it crosses a point $5 \, m$ below the top,another stone starts to fall from a point $25 \, m$ below the top. Both stones reach the bottom of the building simultaneously. The height of the building is ..... $m$.

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