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Q.

A small body is thrown vertically up, which reaches a maximum height of H, and whose total time of flight is T. Its velocity at a height ‘h’ will be                          

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a

8T(Hh)

b

8TH2+Hh

c

4T(H+h)

d

4TH2Hh

answer is A.

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Detailed Solution

Given that

A body is thrown up vertically

Initial velocity = u

Acceleration a=g

Height ascended = h

Final velocity = v

We get

v2u2=2gh

v2=u22gh(1)

Maximum height H=u22g

Time of flight T=2ug

u2=2gH

Now equation (1) will become

v2=2gH2gh

v2=2g(Hh)

H=u22gT=2ug

H=u2g2g2ug=T2

H=g2(ug)2

H=g2(T24)

g=8HT2

v2=2×8HT2(Hh)               

v2=16T2(H2Hh)

v=4TH2Hh

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