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?

physics  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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