An object of mass $10kg$ is allowed to fall from rest from a height of $4.9m$. What will be its velocity when it just touches the ground, if acceleration due to gravity is $9.8m/s^{2}$.

Given:

Mass of the body, $m$ = 10 kg.

Height, $h$ = 4.9 m.

Acceleration due to gravity, $g$ = 9.8 $m/s^2$.

To find: Velocity, $v$ of the object.

Solution:

We know that the formula for potential energy is given as-

$P.E=mgh$

Substituting the given values, we get-

$P.E=10\times {9.8}\times {4.9}$

$P.E=480.2J \sim 480J$

$P.E=480J$

Now, by the law of the conservation of the energy, we know that-

Total amount of energy of the system remains constant.

$\therefore$ Kinetic Energy when the object just touches the ground is equal to the Potential Energy at the height of 4.9 m.

$\therefore$ Kinetic Energy = 480 J.

We know that formula for kinetic energy is given as-

$K.E=\frac {1}{2}\times {mv^2}$

Putting the given values we get-

$480=\frac {1}{2}\times {10v^2}$

$480=5v^2$

$v^2=\frac {480}{5}$

$v^2=96$

$v=\sqrt{96}$

$v=9.79m/s$

Thus, the velocity, $v$ of the object when it just touches the ground is 9.79 m/s.