A person jumps onto a cement …
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Yogita Ingle 5 years ago
h=1 m. h'=9 m.
t=0.1 s. t'=1 s.
Now force exerted in first case be F and in second case be F'.
Let the mass of man be m.
F= ma = m(change in velocity/change in time) = m(v-0)/t = m(sqrt(2gh))/t
F'=m(sqrt(2gh'))/t'
=> F/F'= sqrt(h/h')*t'/t
=>F/F'= sqrt(1/9) *1/0.1 = 10/3
Hence , F'=(3/10)F.
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