Question

A hammer falls freely from the height of 2 m. Calculate the velocity (in m/sec) of the hammer when it just touches the ground. $(takeg=10m/se{c}^2)$

Answer 1

From conservation of energy

We know that just before touching the ground full potential energy will transform into the kinetic energy;

So, $\frac{m{v}^2}{2}$ = mgh;

$\frac{v^2}{2}$ = gh;

$v^2$ = 2gh = 2*10*2

V = $\sqrt{40}$

= 6.32 m/sec