A stone is dropped from the top of a tower of height h. After 1 s, another stone is dropped from a balcony 20 m below the top. Both the stones reach the bottom simultaneously. The value of h is
$[Takeg=10m/s^2]$

The correct option is C
$31.25m$
The stone is dropped from top of the tower (i.e., point A) of height h and it reaches the bottom (i.e., point C) in time t.

After 1 s, another stone is dropped from the balcony, i.e., from point B, which is at height h = 20. It reaches point C in time (t – 1) s.

Initial velocity of stone, u = 0
(As stone is being dropped)


Using second equation of motion for the first stone, $h=ut+\frac{1}{2}g{t}^2\Rightarrow h=0+\frac{1}{2}\times 10t^2(\because g=10m/s^2)\Rightarrow h=5t^2...\left(i\right)$

For second stone (dropped from point B):

$h-20=0+\frac{1}{2}g(t-1{)}^2\text{Using second equation of motion}h-20=5(t-1{)}^2...\left(ii\right)$
Putting the value of h from (i) in (ii), we get:

$5t^2–20=5(t–1{)}^2\Rightarrow 5t^2–20=5(t^2+1–2t)\Rightarrow 5t^2–20=5t^2+5–10t\Rightarrow t=2.5s\text{Using t = 2.5 s in (i), we get:}h=5(2.5{)}^2\Rightarrow h=5\times 6.25\Rightarrow h=31.25m$

Hence, the correct answer is option (c).