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 has reached at a speed of $30m/s$? (Take g= $10m/s^2$)

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

Answer 1

The correct option is D 40 m

For first stone u = 0, v = $30m/s$
$v^2=u^2+2gh$
$(30{)}^2=2\times 10h$
h = 45 m
$45=0+\frac{1}{2}\times 10\times t^2$
$t^2=9$
t = 3 sec
The second stone is dropped after 2 seconds, so by the time the first stone hits the ground , the second one was falling freely for 1 second.
Therefore, t = 1 sec
$S=\frac{1}{2}\times 10\times 1$
$\Rightarrow$s = 5 m
So, they are 45 - 5 = 40 apart from each other.