Question

A stone falls freely under gravity. It covers distance $h_1,h_2$ and $h_3.$ In first $5$ sec, next $5$ sec and next $5$ sec respectively. The relation between $h_1,h_2$ and $h_3$ is

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

Answer 1

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

For a particle released from a certain height the distance covered by the particle in relation with time is given by, $h=\frac{1}{2}g{t}^2$

For first 5 sec ,$h_1=\frac{1}{2}g(5{)}^2=125$

Further next 5 sec $h_1+h_2=h=\frac{1}{2}g(10{)}^2=500$

$\Rightarrow h_2=375$

$h_1+h_2+h_3=\frac{1}{2}g(15{)}^2=1125$

$\Rightarrow h_3=625$

So, $h_2=3h_1$
$h_3=5h_1$

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