Physics - W.E.P. Question with Solution | TestHub

Let the mass of the ball be $m$.

Initial total energy of the ball at height $h_1=30$ m is:

$E_i=mg{h}_1+\frac{1}{2}m{V}_0^2$

After colliding with the ground, the ball loses $60\%$ of its energy. The remaining energy is $40\%$ of the initial energy:

$E_f=0.4E_i$

The ball rebounds to a height $h_2=20$ m, so its final energy at the highest point is:

$E_f=mg{h}_2$

Equating the two expressions for $E_f$:

$0.4\left(mg{h}_1+\frac{1}{2}m{V}_0^2\right)=mg{h}_2$

Substituting $g=10$ m/s${}^2$, $h_1=30$ m, and $h_2=20$ m:

$0.4\left(10\times 30+\frac{1}{2}{V}_0^2\right)=10\times 20$

$0.4\left(300+\frac{1}{2}{V}_0^2\right)=200$

$300+\frac{1}{2}{V}_0^2=\frac{200}{0.4}$

$300+\frac{1}{2}{V}_0^2=500$

$\frac{1}{2}{V}_0^2=200$

$V_0^2=400\Rightarrow V_0=20$ m/s

Answer: $20$ m/s