Question

A rock is dropped from the top of a tall building. The rock’s displacement in the last second before it hits the ground is 45% of the entire distance it falls. How tall is the building?

Short answer

The height of the building is 73.7 m

Step-by-step solution

Step 1. Write the given information. 

The initial velocity of the rock is $v_0$
The time at this instant is $t_0$
A second before hitting the ground, the distance traveled by rock is $h_1$
The velocity at this point is $v_1$
The time taken by a rock to reach this point is $t_1$

The time taken by the rock to reach the ground is $t_2$

The height of the building is H.

It is given that $t_2-t_1=1sec$and, $h_1=45\%ofH$

localid="1648190095951" $$\begin{gathered} \Rightarrow h_1=\frac{45}{100}H \\ {h}_1=\frac{9}{20}H \end{gathered}$$

Step 2. To determine the time taken by a rock to reach points B and C 

When the rock reaches the point B from point A,
Using the equation of motion
$$\begin{gathered} H-h_1=v_{0}t+\frac{1}{2}g{t_1}^2 \\ H-\frac{9}{20}H=0+\frac{1}{2}g{t_1}^2 \\ \frac{11}{20}H=\frac{1}{2}g{t_1}^2 \\ {t_1}^2=\frac{22H}{20g} \\ {t}_1=\sqrt{\frac{22H}{20g}}.....\left(1\right) \end{gathered}$$

Similarly, when the rock reaches point C,

Using the equation of motion
$$\begin{gathered} H-h_2=v_0{t}_2+\frac{1}{2}g{t_2}^2 \\ H-0=0+\frac{1}{2}g{t_2}^2 \\ \frac{1}{2}g{t_2}^2=H \\ {t_2}^2=\frac{2H}{g} \\ {t}_2=\sqrt{\frac{2H}{g}}......\left(2\right) \end{gathered}$$

Step 3. To determine the height of the building.

As it is given in that $t_2-t_1=1sec$
Therefore, take the difference between equations (1) and (2)
localid="1648191195111" $$\begin{gathered} \sqrt{\frac{2H}{g}}-\sqrt{\frac{22H}{20g}}=1 \\ \sqrt{\frac{2H}{g}}\left(1-\sqrt{\frac{11}{20}}\right)=1 \\ \sqrt{\frac{2H}{g}}={\left(1-\sqrt{\frac{11}{20}}\right)}^{-1}={\left(1-0.74\right)}^{-1}={\left(0.26\right)}^{-1} \\ \sqrt{\frac{2H}{g}}=3.84 \\ squaringbothsides \\ \frac{2H}{g}=14.7 \\ H=\frac{14.7\left(10\right)}{2}=73.7m \end{gathered}$$
Hence, the height of the building is 73.7 m