Question

A ball dropped from a height of 10m looses $40\%$ of energy after striking the ground. How much high can the ball bounce back? [ g= 10]

Answer 1

$H_1=10m$, $v_1=?$
Loss of energy $=40\%$
$H_2=?$
On striking the floor , K.E = P.E
$\frac{1}{2}m{v}^2=mg$
$H_1{V}_1=?$
$\left(2g{H}_1\right)=?$
\((2 \times 9.8 \times 10)V_1=?\)
$196=14m/s$
As the body loses $40\%$ of its energy , final energy $E_2=\left(\frac{60}{100}\right)$ of initial energy i.e. $mg{H}_2=\left(\frac{60}{100}\right)mg{H}_1$
\(H_2=\frac{3}{5} H_1=\left ( \frac{3}{5} \right ) \times 10 = 6~m\)