A ball of mass 50gm is thrown upwards. It rise to a height of 100m. At what height kinetic energy is reduced to 70percent?
Asked by ravinecw | 21st Nov, 2018, 10:32: AM
Initial speed of projection is obtained from the relation v2 = u2 - 2 g h ....................(1)
where v is final speed and u is initial speed and h is the height reached by the ball when its speed is v.
At maximum height v will be zero.
hence initial speed u is obatined from : u2 = 2gh = 2×9.8×100 = 1960
when the kinetic energy is 70% of initial kinetic energy means, in eqn.(1) we substitute v2 = 0.7u2
Hence using this substitution, eqn .(1) can be modified as : h' = 0.3u2 / (2g) .....................(2)
where h' is the heiight reached by the ball when its kinetic energy is 70% of its initial kinetic energy
h' = 0.3×1960/19.6 = 30 m
Answered by Thiyagarajan K | 21st Nov, 2018, 11:58: AM
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