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Question

# A ball is thrown vertically upwards. It was observed at a height h, twice with a time interval Δt . The initial velocity of the ball is

A
8gh+g2(Δt)2
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B
8gh+(gΔt2)2
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C
128gh+g2(Δt)2
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D
8gh+4g2(Δt)2
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Solution

## The correct option is D 12√8gh+g2(Δt)2If the ball is reaching height h then, displacement=hTime taken=t1... (say)Acceleration=−gInitial velocity=u1∴from the equations of motionS=ut+12at2∴h=u1t1+12(−g)t21.....(i)If the ball is reaching height 2hthen, displacement=2hTime taken=t1+△tAcceleration=−gInitial velocity=u2Therefore from equations of motion2h=u2(t1+△t)−12g(t1+△t)2.....(ii)Also we know from the equations of motion that v=u+at∴0=u1−gt1....(iii)& 0=u2−g(t1+△t).....(iv)∴u1=gt1,u2=gt1+g△t=u1+g△t∴From equations (i),(ii),(iii),(iv)we get(ii)−(i)=h={(u1+g△t)(t1+△t)−12g(t1+△t)2−u1t1−12gt21}∴u1=12√8gh+g2(△t)2 Thus option C

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