Question

A stone falls from a building and reaches the ground 2.5 seconds later. How high is the building? ( $g = 9.8m/{s^2}$ )

Answer 1

We can use the second equation of motion to solve the above problem. The stone is said to fall, but with no initial velocity specified, hence assume initial velocity to be equal to zero.
Formula used: In this solution we will be using the following formulae;
$h = ut + \dfrac{1}{2}g{t^2}$
where $h$ is the vertical distance travelled by an object falling under gravity, $u$ is the initial velocity (vertical) the body was projected with, $g$ is the acceleration due to gravity, and $t$ is the time taken to fall the vertical distance

Complete Step-by-Step solution:
To calculate the value of the height of the building using the given values, we note that the initial velocity must be zero. Generally, we can use the second equation of motion, which can be given by the equation,
$h = ut + \dfrac{1}{2}g{t^2}$
where $h$ is the vertical distance travelled by an object falling under gravity, $u$ is the initial velocity (vertical) the body was projected with, $g$ is the acceleration due to gravity, and $t$ is the time taken to fall the vertical distance
Hence, by inserting known values, we have $h = \left( 0 \right)\left( {2.5} \right) + \dfrac{1}{2}\left( {9.8} \right){\left( {2.5} \right)^2}$
By calculating, we have that
$h = \dfrac{1}{2}\left( {9.8} \right)6.25$
$\Rightarrow h = 30.625m$
or approximately 31 m.

Note: For clarity, note that this equation is only accurate if the air resistance is negligible. In the presence of air resistance, the equation overestimates the height of the building. This is because it would take a longer period for the stone to get to the ground with air resistance presence.
In any case, the height is only correct if the air resistance on the stone is negligible.