$A$ stone is dropped from the edge of a roof.
$(a)$ How long does it take to fall $4.9 \, m$?
$(b)$ How fast does it move at the end of that fall?
$(c)$ How fast does it move at the end of $7.9 \, m$?
$(d)$ What is its acceleration after $1 \, s$ and after $2 \, s$?

(N/A) Given: Initial velocity,$u = 0 \, m/s$.
Acceleration due to gravity,$g = 9.8 \, m/s^2$.
$(a)$ Using the equation of motion $S = ut + \frac{1}{2}at^2$:
$4.9 = 0 \times t + \frac{1}{2} \times 9.8 \times t^2$
$4.9 = 4.9 \times t^2$
$t^2 = 1 \Rightarrow t = 1 \, s$.
So,it takes $1 \, s$ to fall $4.9 \, m$.
$(b)$ Using $v^2 - u^2 = 2aS$:
$v^2 - 0^2 = 2 \times 9.8 \times 4.9$
$v^2 = 96.04$
$v = 9.8 \, m/s$.
So,the speed at the end of $4.9 \, m$ is $9.8 \, m/s$.
$(c)$ Using $v^2 - u^2 = 2aS$:
$v^2 - 0^2 = 2 \times 9.8 \times 7.9$
$v^2 = 154.84$
$v = \sqrt{154.84} \approx 12.44 \, m/s$.
So,the speed at the end of $7.9 \, m$ is $12.44 \, m/s$.
$(d)$ During free fall,the acceleration due to gravity is constant.
Acceleration after $1 \, s = 9.8 \, m/s^2$.
Acceleration after $2 \, s = 9.8 \, m/s^2$.