Question

A stone falls freely under gravity. It covers distances $h_1,h_2$ and $h_3$ in the first $5$ seconds, the next $5$ second and the next $5$ seconds respectively. The relation between $h_1,h_2$ and $h_3$ is-

• A. $h_1=h_2=h_3$
• B. $h_1=2h_2=3h_3$
• C. $h_1=\frac{h_2}{3}=\frac{h_3}{5}$
• D. $h_2=3h_1\text{and}{h}_3=3h_2$

Answer 1

The correct option is C $h_1=\frac{h_2}{3}=\frac{h_3}{5}$

Using, first kinematic equation of motion,

$S=ut+\frac{1}{2}a\left(t{)}^2\right[\because u=0,a=g]$

For the interval $t=0\text{to}t=5$, $h_1=\frac{1}{2}g(5{)}^2=125\text{m}$

For the interval $t=0\text{to}t=10$
$h_1+h_2=\frac{1}{2}g(10{)}^2=500\text{m}$

$\therefore h_2=500-125=375\text{m}$

For the interval $t=0\text{to}t=15$
$h_1+h_2+h_3=\frac{1}{2}g(15{)}^2=1125\text{m}$

$\therefore h_3=1125-500=625\text{m}$

$\therefore h_1=\frac{h_2}{3}=\frac{h_3}{5}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.