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April 25, 2015

April 25, 2015

Posted by **Edwin** on Sunday, March 28, 2010 at 2:59pm.

- physics -
**Afolabi Emmanuel Ayo**, Sunday, April 15, 2012 at 7:04amthe acceleration as it's coming down is +10. Let the velocity at which it hit the ground be V. So, the velocity at which it rebounds will be 3/4V. Thus, 4rm v=u+gt. Where u is 0, v=10t, d time at which it lands is t=v/10. 4rm s=(v-u)/2 *t, 20=(v-3/4v)/2 *v/10. Thus, 400=v^2/4. V^2=1600, v=40. Now, d time for hitting d ground is t=40/10, t=4s. 4rm v=u+(-g)t, the time for rebounce t=30-40/-10, t=1s. Thus, d time interval btw the first & second bounce is t1-t2=4-1=3s