A body falling from a height of 20m rebounds from hard floor. If it loses 20% energy in the impact, then coefficient of restitution is

The correct option is 0.89.

Step-by-step Explanation:

Let the mass of the body falling from the height be $m$.

When the body falls from a height $h_1=20\text{m}$, its potential energy is converted completely into kinetic energy just before it strikes the hard floor.

The kinetic energy of the body just before the impact ($K_i$) with velocity $u$ is given by:
$K_i=\frac{1}{2}m{u}^2$

During the impact, the body loses 20% of its kinetic energy. Therefore, the remaining kinetic energy just after the impact ($K_f$) with rebound velocity $v$ is 80% of the initial kinetic energy:
$K_f=80\%\text{of}{K}_i$
$K_f=0.80\times K_i$

Substituting the expressions for kinetic energy in terms of velocities:
$\frac{1}{2}m{v}^2=0.80\times \left(\frac{1}{2}m{u}^2\right)$

Simplifying the equation by canceling $\frac{1}{2}m$ from both sides:
$v^2=0.80u^2$

Taking the square root on both sides, we get:
$\frac{v}{u}=\sqrt{0.80}$

The coefficient of restitution ($e$) is defined as the ratio of the velocity of separation to the velocity of approach:
$e=\frac{v}{u}$

Thus, we have:
$e=\sqrt{0.80}\approx 0.894$

Rounding off to two decimal places, we get:
$e\approx 0.89$