Physics of Bouncing: Calculating Velocity & Energy Conservation Explor

Step 1 of 3

Consider the given data as follows.

The mass of the ball is m=0.150 kg.

The height from the ball was dropped is \(h{}_{1}=1.25\ \text{m}\)

It rises from the floor to a height of \({{h}_{2}}=0.960\ \text{m}\).

Let the impulse is \(\Delta P\).
For the given system, the kinetic energy is equal to the potential energy. Therefore,

\(\text{Kinetic Energy=Potential Energy}\)

 \(  \dfrac{\text{1}}{\text{2}}\text{m}{{\text{v}}^{\text{2}}}\text{ =m }\!\!\times\!\!\text{ g }\!\!\times\!\!\text{ h} \) 

\(  \dfrac{\text{1}}{\text{2}}{{\text{v}}^{\text{2}}}\text{ =g }\!\!\times\!\!\text{ h} \) 

\( {{\text{v}}^{\text{2}}}\text{ =2 }\!\!\times\!\!\text{ g }\!\!\times\!\!\text{ h} \) 

\(  \text{v =}\sqrt{\text{2 }\!\!\times\!\!\text{ g }\!\!\times\!\!\text{ h}}\ \ ............(1) \)