Question

A ball ( initially at rest ) is released from the top of a tower. The ratio of work done by the force of gravity in the first, second and third seconds is

• A. $1:3:5$
• B. $1:4:9$
• C. $1:9:25$
• D. $1:2:3$

Answer 1

The correct option is A $1:3:5$

$S=ut+\frac{1}{2}a{t}^2$

$u=0$

$H=\frac{1}{2}g{t}^2$

Now H for first second will be

$H=\frac{1}{2}g\left(1\right)$

H for second second will be

$H=\frac{1}{2}g\left(4\right)$

H for third second will be

$H=\frac{1}{2}g\left(9\right)$

Now as we have to Know the distance of second second we need to substract the distance of first second.

So required $H1=g\frac{1}{2}$

$H2=g\frac{4}{2}-g\frac{1}{2}=g\frac{3}{2}$

$H3=g\frac{9}{2}-g\frac{4}{2}=g\frac{5}{2}$

Put the values of H in$W=mgH$

On dividing above 3eqs we get : $1:3:5$

So required ratio is $1:3:5$