Question

Two bodies fall freely from different heights and reach the ground simultaneously. The time of descent for the first body is 1s and second body is 2s. At what height was the first body situated when the other began to fall?

Answer 1

The initial height of the first body is given by:
$h_1 = \frac{1}{2}gt^2$
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
$h_1 = \frac{1}{2}(9.81 m/s^2)(1 s)^2=4.9 m$

The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
$h_2 = \frac{1}{2}(9.81 m/s^2)(2 s)^2=19.6 m$

The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.