Question

A stone is allowed to fall from the top of a tower $100\text{m}$ high and at the same time another stone is projected vertically upwards from the ground with a velocity of $25{\text{ms}}^{-1}$. The two stones will meet after

• A. $4\text{s}$
• B. $0.4\text{s}$
• C. $0.04\text{s}$
• D. $2\text{s}$

Answer 1

The correct option is A $4\text{s}$

Using kinematics
Let the distance from top of the tower where the stones meet be $x$,
$x=\frac{1}{2}g{t}^2,100-x=25t-\frac{1}{2}g{t}^2$
Adding both the equations we get
$25t=100$ or $t=4\text{s}$

Using Relative velocity:

Initial velocity of stone 1 which is dropped from the top of a tower $=u_1=0$
Initial velocity of stone 2 which is thrown upward from the ground $=u_2=+25\text{m/s}$
(Take upward direction to be positive)
Hence, $u_{rel}=u_{21}=u_2-u_1=25-0=25\text{m/s}$
$S_{rel}=100\text{m}$
$a_{rel}=g-g=0$
Using second equation of motion, we get
$S_{rel}=u_{rel}t+\frac{1}{2}{a}_{rel}{t}^2$
$\Rightarrow 100=25\times t+0$
$\therefore t=4\text{s}$
Hence, the two stones will meet after 4 seconds.