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

Expert Answer:

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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