The distance travelled by a body during last second of its total flight is d when the body is projected vertically up with certain velocity. If the velocity of projection is doubled, The distance travelled by the body during last second of its total flight is

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$2 d + \frac{g}{2}$

$2 d - \frac{g}{2}$

answer is D.

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Detailed Solution

$s_{n} = u + a [n - \frac{1}{2}]$

$= u - g [\frac{2 u}{g} - \frac{1}{2}]$

$= u - 2 u + \frac{g}{2}$

$d = - u + \frac{g}{2}$

$d^{1} = 2 u - g [\frac{4 u}{g} - \frac{1}{2}] = - 2 u + \frac{g}{2}$

$= - 2 u + g - \frac{g}{2}$

$= 2 [- u + \frac{g}{2}] - \frac{g}{2}$

$= 2. d - \frac{g}{2}$

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