Question

A stone is dropped by a person from the top of a tower, which is 200 m tall. At the same time another stone is thrown upwards, with a velocity of $50{ms}^{-1}$ by a person standing at the foot of the tower. Find the time after which two stones meet.

• A. $4s$
• B. $5s$
• C. $8s$
• D. $10s$

Answer 1

The correct option is A $4s$

Let the two stones meet at a distance of x mfrom the top of the tower, and 't' be the time taken. Let us assume the downward direction as positive.
For the stone that is dropped its initial velocity $u=0{ms}^{-1}$ ; displacement s = x and acceleration = acceleration due to gravity (g).
Using $s=ut+\frac{1}{2}{at}^2,wegetx=\left(0\right)t+\frac{1}{2}{gt}^2\to \left(1\right)$.
For the stone that is projected vertically upwards, its initial velocity, $u=-50{ms}^{-1}$; displacement s =-(200-x) and acceleration a =g.
using $s=ut+\frac{1}{2}{at}^2,weget-(200-x)=-50\times t+1/2{gt}^{2}200=50t-1/2{gt}^2+x\to \left(1\right)\left(2\right)Fromtheequations\left(1\right)and\left(2\right)wehave200=50t-1/2{gt}^2+1/2{gt}^2\Rightarrow 200=50\therefore t=4s$