Posted by **sakshi** on Friday, June 15, 2012 at 9:11am.

a body is thrown vetically up from the ground passes the height 10.2 m twice at an interval of 10s .what was initial velocity?

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**drwls**, Friday, June 15, 2012 at 11:04am
The body spent 5.1 s going up above 10.2 m, and another 5.1 s going down. Its speed at that point was therefore

v = 9.8*5.1 = 49.98 m/s @ y = 10.2 m_

For the initial velocity Vo, use conservation of energy:

Vo^2/2 - g*10.2 m = v^2/2

Vo^2/2 = 1249 + 100

Vo = 51.9 m/s

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