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Q. A ball which is at rest is dropped from a height $ h $ metre. As it bounces off the floor its speed is $ 80% $ of what it was just before touching the ground. The ball will then rise to nearly a height :

KEAMKEAM 2004

Solution:

$ v=\sqrt{2gh} $ ...(1)
After rebounce, $ {{v}^{2}}={{u}^{2}}-2gh $
$ \Rightarrow $ $ {{u}^{2}}={{v}^{2}}+2gh $
$ \therefore $ $ {{u}^{2}}=2gh' $ ...(2) $ \therefore $
$ \frac{{{v}^{2}}}{{{u}^{2}}}=\frac{2gh}{2gh} $
$ \Rightarrow $ $ h'=h\times \frac{{{u}^{2}}}{{{v}^{2}}}=h\times {{\left( \frac{80}{100} \right)}^{2}} $
$ =0.64\,h $