A stone dropped from the top of a tower travels 4.9 m in the last second, then the velocity of the stone on reaching the ground is

19.6 m^{- 1}

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9.8 m s^{- 1}

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4.9 m s^{- 1}

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29.4 m s^{- 1}

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Solution

The correct option is C $9.8 m s^{- 1}$
Using the formula for the distance travelled in $n^{t h}$ second.
$s_{n} = u + \frac{a}{2} (2 n - 1)$
$4.9 = \frac{g}{2} (2 n - 1)$
$n = 1 s$
$n^{t h}$ second is the last second which is $1^{s t}$ second.
$v = u + a t$
$v = g \times 1 = 9.8 m / s^{2}$

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A stone dropped from the top of a tower travels 4.9 m in the last second, then the velocity of the stone on reaching the ground is

The correct option is C 9.8ms−1Using the formula for the distance travelled in nth second.sn=u+a2(2n−1)4.9=g2(2n−1)n=1 snth second is the last second which is 1st second.v=u+atv=g×1=9.8 m/s2

The correct option is C $9.8 m s^{- 1}$
Using the formula for the distance travelled in $n^{t h}$ second.
$s_{n} = u + \frac{a}{2} (2 n - 1)$
$4.9 = \frac{g}{2} (2 n - 1)$
$n = 1 s$
$n^{t h}$ second is the last second which is $1^{s t}$ second.
$v = u + a t$
$v = g \times 1 = 9.8 m / s^{2}$