91Ó°ÊÓ

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

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,

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

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

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

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

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

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

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