Question

A stone is dropped from the roof of a high building. a second stone is
dropped 1 sec later. How far apart are the stone when the second one
has reached a speed of 23 m/s?​

Answer 1

Explanation:

The key is to find in how much time will the second stone reach 12 m/s.

v = u + at, here a = 9.8 m/s2 (accl due to gravity).

12 = 0 + 10 * t => t = 1.2 sec

distance covered by 2nd stone in 1.2 sec = 1/2 gt^2 = 1/2 *10 * (12/10)^2 = 7.2 m

Now, the time for first stone to travel down = 1.5 + 1.2 = 2.7 sec

distance covered by 1st stone in 2.7 = 1/2 * 10 * 2.7^2 = 36.5 m

Hence distance b/w the two = 36.5 - 7.2 = 29.3 m