If a particle is thrown with velocity more than $10 m / s$ vertically upward, then the distance traveled by the particle in last second of its ascent is:

g

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\frac{g}{2}

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\frac{g}{4}

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\frac{g}{8}

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Solution

The correct option is B $\frac{g}{2}$
time of ascent be t
$0 = 10 - g t$
$t = 1 s$
$s = u t - \frac{1}{2} g t^{2} = \frac{g}{2}$

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