a body of mass 3 kg is dropped from a tower of height 250m,then its kinetic energy after 3s will be????
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Given that , m = 3kg , u = 0 , t = 3 sec , a = g = 9.8m/s2
So velocity at t = 3sec
v = u +a t
= 0 + 9.8(3)
= 29.4 m/s
Kinetic energy = 1/2 m v2
= 1/2 (3) (29.4)2
= 1296.54 J
So velocity at t = 3sec
v = u +a t
= 0 + 9.8(3)
= 29.4 m/s
Kinetic energy = 1/2 m v2
= 1/2 (3) (29.4)2
= 1296.54 J
maithilimundle:
thank you sooooooooooo much
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