Question

A rock is dropped from a $\mathbf{100}\mathbf{}\mathit{m}$high cliff. How long does it take to fall (a) the first $\mathbf{50}\mathbf{}\mathit{m}$and (b) the second$\mathbf{50}\mathbf{}\mathit{m}$?

Short answer

1. The time taken for the ball to fall for the first $50\mathrm{m}$is$3.2\mathrm{s}$
2. The time taken for the ball to fall for the second $50\mathrm{m}$isrole="math" localid="1663068485089" $1.3\mathrm{s}$

Step-by-step solution

Given information

$\begin{array}{l}{y}_1=50\mathrm{m}\\ y_2=100\mathrm{m}\end{array}$

understand the concept

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 kinematic equation is given by,

$y_1=v_0{t}_1+\frac{1}{2}a{t}_1^2$

(a): Calculations for the time taken by first ball to cover 50 m 

To find time for$50\mathrm{m}$use following equation

$\begin{array}{rcl}{y}_1& =& v_0{t}_1+\frac{1}{2}a{t}_1^2\\ 50& =& 0\times t_1+\frac{1}{2}\times 9.81\times t_1^2\\ t_1& =& 3.2\sec \end{array}$

It takes $3.2\sec$to fall through the first$50\mathrm{m}$distance.

(b): Calculations for the time taken by second ball to cover 50 m

Now to find time for the second$50\mathrm{m}$

$\begin{array}{rcl}{y}_2& =& v_i{t}_2+\frac{1}{2}a{t}_2^2\\ 100& =& 0\times t_1+\frac{1}{2}\times 9.81\times t_2^2\\ t_2& =& 4.5\sec \end{array}$

To fall through the total of$100\mathrm{m}$it takes$4.5\sec$.

So, the time for the second $50\mathrm{m}$is

role="math" localid="1663069535266" $\begin{array}{c}t=t_2-t_1\\ =4.5-3.2\\ =1.3\sec .\end{array}$