Question

A stone is dropped from a building and $2$ seconds later another stone is dropped. How far apart are these two stones by the time the first one reaches a speed of $30m{s}^{-1}$? (Take $g=10m{s}^{-2}$)

• A. $80m$
• B. $100m$
• C. $60m$
• D. $40m$

Answer 1

The correct option is D $40m$

$v=u+atorv=9.8t$

$t=\frac{30}{10}=3.0s$ is the time at which stone will reach 30 m/s velocity

First stone will reach $s=ut+\left(1/2\right)a{t}^2$

so $s_1=1/2\times 10\times 9m$

$s_1=45m$

(initial velocity = u = 0)

second stone was dropped after $2sec$, so time for it$t_1=3.0-2.0s=1.0s$

$s_2=1/2\times 10\times 1^2=5.0m$

so distance between two stones

$D=45.0-5.0=40.400m$

Option D is correct.