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

The impact between the ball and the plate is direct impact. If u is the velocity of the ball before impact, its velocity after is eu, where e is the coefficient of restitution.

But$u=\sqrt{2\text{gh}}$since the ball falls from a height h. So the velocity of the ball at the first rebound

$={\text{v}}_1=eu=e\sqrt{2\text{gh}}$

The height h₁to which it rises after first rebound

$h_1=\frac{{\text{v}}_1^2}{2\text{g}}=\frac{e^2·2\text{gh}}{2\text{g}}=e^{2}h$

The velocity v₂after second rebound v₂= ev₁= e²u

The height h₂it rises after second rebound

$h_2=\frac{{\text{v}}_2^2}{2\text{g}}=\frac{e^4·u^2}{2\text{g}}=\frac{e^4·2\text{gh}}{2\text{g}}=e^{4}h$

In general, the height hn to which the ball rises after n^threbound is given by h_n= e²ⁿh.

Total distance travelled

= h + 2h₁+ 2h₂+ 2h₃+ ................+2h_n+............to ∞

= h + 2 (e²h + e⁴h + e⁶h +.......)

= h + 2he²(1 + e²+ e⁴+...........)

$=h+\frac{2h{e}^2}{1-e^2}=h\left(1+\frac{2e^2}{1-e^2}\right)=\frac{h\left(1+e^2\right)}{\left(1-e^2\right)}$

$=\frac{9\left(1+{\left(\frac{1}{2}\right)}^2\right)}{\left(1-{\left(\frac{1}{2}\right)}^2\right)}=15\text{m}$