Today I came across the question below. I thought to write about it explaining how to find the total distance traveled by a bouncing object. Photo by cottonbro: Question: A particle falls from h height and bounces. Prove that the total distance traveled is \(\frac{\left ( 1+e^{2} \right )}{\left ( 1-e^{2} \right )}h \). and the total time is \( \sqrt{\frac{2h}{g}} \cdot \frac{(1+e)}{1-e}\) where e is the elasticity coefficient. we can find the particle's velocity when it hits the ground for the first time using the law of mechanical energy conservation as shown below. \(mgh = \frac{1}{2}mv^{2}\) \(\Rightarrow v = \sqrt{2gh}\) and we can use Newton's law to find out the velocity (v) of the particle when it bounces for the first time. \(e = \frac{v}{\sqrt{2gh}}\) \(\Rightarrow v = e\sqrt{2gh} \) According to this, we can determine the particle's speeds when it hits the ground are in a geometric series as follows. \( e\sqrt{2gh}, e^{2}\sqrt{2gh}, e^3\sqrt{2gh},....\) The n