Question: A rock is dropped (from rest) from the top of atall building. How far above the ground is the rock$\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{s}$before it reaches the ground?

${\mathrm{y}}_0=60\mathrm{m}$

In this case

${\mathrm{v}}_0=0$

y= position of rock above the ground after t second

The problem deals with the kinematic equation of motion. Kinematics is the study of how a system of bodies moves without taking into account the forces or potential fields that influence the motion. The equations which are used in the study are known as kinematic equations of motion.

Formula:

The displacement in the kinematic equation under free fall acceleration is given by,

$\mathrm{y}-{\mathrm{y}}_0={\mathrm{v}}_0\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2$………………….(i)

Using equation (i),

$$\begin{gathered} \mathrm{y}-{\mathrm{y}}_0={\mathrm{v}}_0\times \mathrm{t}-\frac{1}{2}\times \mathrm{g}\times {\mathrm{t}}^2 \\ \Rightarrow 0-60=0\times \mathrm{t}-\frac{1}{2}\times 9.81\times {\mathrm{t}}^2 \\ \Rightarrow \mathrm{t}=3.5\mathrm{s} \end{gathered}$$

So, the total time taken by the stone to hit the ground is $3.5\mathrm{s}$.

As we have to find the position at $1.2\mathrm{s}$earlier which means at $3.5-1.2=2.3\mathrm{s}$

$$\begin{gathered} \mathrm{y}-{\mathrm{y}}_0={\mathrm{v}}_0\times \mathrm{t}-\frac{1}{2}\times \mathrm{g}\times {\mathrm{t}}^2 \\ \Rightarrow y-60=0\times 2.3-\frac{1}{2}\times 9.81\times 2.3^2 \\ \Rightarrow \mathrm{y}=34.01\mathrm{m} \end{gathered}$$

So,

$\mathrm{y}=34.01\mathrm{m}$

The distance of the rock at $1.2\mathrm{s}$before it reaches the ground is $34.01\mathrm{m}.$