al setup: The ball falls freely from rest, with a velocity just before the first bounce of $7 \, \mathrm{ms}^{-1}$, and the coefficient of restitution $e = 0.75$. The acceleration due to gravity is $10 \, \mathrm{ms}^{-2}$.

Distance traveled: The ball falls a height $h$ before the first bounce. After each bounce, it reaches successively lower heights $h_1, h_2, \ldots$, where:

$h_1 = e^2 h$

$h_2 = e^4 h$

And so on.

Total distance formula: The total distance $d$ covered by the ball is:

$ d = h + 2h_1 + 2h_2 + \ldots $

This can be rewritten using geometric series:

$ d = h + 2e^2h (1 + e^2 + e^4 + \ldots) $

Using the sum of an infinite geometric series $S = \frac{1}{1 - e^2}$, we have:

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

Simplifying, we get:

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

Plugging in the values: Given $e = \frac{3}{4}$, the distance $d$ becomes:

$ d = h \left(\frac{1 + \left(\frac{3}{4}\right)^2}{1 - \left(\frac{3}{4}\right)^2}\right) = \frac{25}{7} \times h $

Calculating $h$:

Using the kinematic equation:

$ v^2 = u^2 + 2gh \quad \Rightarrow \quad (7)^2 = 0 + 2 \times 10 \times h $

Solving for $h$, we find:

$ \frac{49}{20} = h $

Finding total distance $d$:

Using the relation obtained:

$ d = \frac{25}{7} \times \frac{49}{20} = 8.75 \, \mathrm{m} $

Therefore, the total distance traveled by the ball before it comes to rest is $8.75 \, \mathrm{m}$.

Center Of Mass

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