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(D) Let the two stones meet after a time $t$.For the stone dropped from the tower:Initial velocity,$u = 0\, m/s$.Let the displacement of the stone in

Vedclass · Accepted Answer

$78.4\, m$ and $20.6\, m$

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(D) Let the two stones meet after a time $t$.For the stone dropped from the tower:Initial velocity,$u = 0\, m/s$.Let the displacement of the stone in time $t$ from the top of the tower be $s$.Acceleration due to gravity,$g = 9.8\, m/s^2$.Using the equation of motion,$s = ut + \frac{1}{2}at^2$:$s = 0 	imes t + \frac{1}{2} 	imes 9.8 	imes t^2$$s = 4.9t^2$ ............. $(1)$For the stone thrown upwards:Initial velocity,$u = 25\, m/s$.Let the displacement of the stone from the ground in time $t$ be $s'$.Acceleration due to gravity,$g = -9.8\, m/s^2$.Using the equation of motion,$s' = ut + \frac{1}{2}at^2$:$s' = 25t - \frac{1}{2} 	imes 9.8 	imes t^2$$s' = 25t - 4.9t^2$ ............. $(2)$The sum of the displacements of both stones at the meeting point is equal to the total height of the tower,$100\, m$:$s + s' = 100$$4.9t^2 + 25t - 4.9t^2 = 100$$25t = 100$$t = 4\, s$In $4\, s$,the distance covered by the falling stone from the top is:$s = 4.9 	imes (4)^2 = 4.9 	imes 16 = 78.4\, m$.Therefore,the stones will meet after $4\, s$ at a height of $(100 - 78.4) = 20.6\, m$ from the ground.

$A$ stone is allowed to fall from the top of a tower $100\, m$ high and at the same time another stone is projected vertically upwards from the ground with a velocity of $25\, m/s$. Calculate when and where the two stones will meet.

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