A steel ball is dropped from a building’s roof and passes a window, taking 0.125 sto fall from the top to the bottom of the window, a distance of 1.20 m. It then falls to a sidewalk and bounces back past the window, moving from bottom to top in 0.125 s. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is $\mathbf{\text{2.00s}}$. How tall is the building?

Time to cover the distance across the window, $∆\mathrm{t}=0.125\mathrm{s}$

Height of the window ${\mathrm{y}}_2-{\mathrm{y}}_1=1.20\mathrm{m}$

Time taken by the ball to travel from the bottom of the window to the floor and back to the bottom of the window is 2 s

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Also it involves the concept of free fall. Using kinematic equations and given height of the window, time taken by ball to travel across the window, the velocity of the ball when it was at the top of the window can be found. Using this velocity, it is possible to find the height of the building above the top of the window. Using the velocity at the top of the window as our initial velocity, top of the window as starting point, and time taken by the ball to reach the bottom from the top of the window, the height of the building can be calculated.

From the given data we can conclude that the time taken by the ball to travel from top of the window to the bottom of building is

$$\begin{gathered} ∆\mathrm{t}=1+0.125\mathrm{s} \\ =1.125\mathrm{s} \end{gathered}$$

$$\begin{gathered} ∆\mathrm{y}={\mathrm{V}}_0∆\mathrm{t}+\frac{1}{2}\mathrm{a}{\left(∆\mathrm{t}\right)}^2 \\ \left(-1.20\right)={\mathrm{V}}_0\left(0.125\right)-\frac{1}{2}\left(9.8\right){\left(0.125\right)}^2 \\ {\mathrm{V}}_0=-8.99\mathrm{m}/\mathrm{s} \end{gathered}$$

We know that the velocity at top position when ball starts falling is zero.

The velocity at top of window is

${\text{v}}_{\text{f}}^2={\text{v}}_0^2+2\text{a}\left({\text{y}}_1-0\right)$

By using third kinematic equation,
$$\begin{gathered} ∆\mathrm{y}={\mathrm{V}}_0∆\mathrm{t}+\frac{1}{2}\mathrm{a}{\left(∆\mathrm{t}\right)}^2 \\ \left({\mathrm{y}}_3-{\mathrm{y}}_1\right)={\mathrm{V}}_0\left(∆\mathrm{t}\text{'}\right)+\frac{1}{2}\mathrm{g}{\left(∆\mathrm{t}\text{'}\right)}^2 \\ \left({\mathrm{y}}_3-\left(-4.12\right)\right)=\left(-8.99\right)\left(1.125\right)-\frac{1}{2}\left(9.8\right){\left(1.125\right)}^2 \\ {\mathrm{y}}_3=-20.4\mathrm{m} \end{gathered}$$

So the height of the building is -20.4 m