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A stone thrown vertically upwards from the surface of the moon at velocity of $24 m / s e c$ . reaches a height of $s = 24 t - 0.8 t 2$ m after t sec. The acceleration due to gravity in $m / \sec^{2}$ at the surface of the moon is:

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$0.8$

$1.6$

$2.4$

$4.9$

answer is B.

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Detailed Solution

$\frac{d s}{d t} =$ Velocity

$= 24 = 24 - 1.6 t$

So, acceleration at $t$ is $[{\frac{d^{2} s}{d t^{2}}|}_{j} = - 1.6$

As stone is thrown upwards, so acceleration due to gravity (which acts downwards)= 1.6.

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