A body is released from the top of a tower of height H m. After 2 s it is stopped and then instantaneously released. What will be its height after next 2 s?

( H - 5) m

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( H - 10) m

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( H - 20) m

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( H - 40) m

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Solution

The correct option is C ( H - 40) m
$v_{f}$ (after $2 s$ ) $= g \times 2 = 20 m / s$
$s_{1} = \frac{1}{2} g t^{2}$
$s_{1} = 20 m$
As it is again released and after $2 s$ it will cover more $20 m$ distance as remaining height $= (H - 40) m$

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A body is released from the top of a tower of height H m. After 2 s it is stopped and then instantaneously released. What will be its height after next 2 s?

The correct option is C ( H - 40) mvf(after 2s)=g×2=20m/ss1=12gt2s1=20mAs it is again released and after 2s it will cover more 20m distance as remaining height=(H−40)m

The correct option is C ( H - 40) m
$v_{f}$ (after $2 s$ ) $= g \times 2 = 20 m / s$
$s_{1} = \frac{1}{2} g t^{2}$
$s_{1} = 20 m$
As it is again released and after $2 s$ it will cover more $20 m$ distance as remaining height $= (H - 40) m$