An object is dropped from a height of $5 m$ and it takes $5 s$ to reach the ground. Calculate the acceleration due to gravity at that place.

0.1 m s^{- 2}

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0.2 m s^{- 2}

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0.3 m s^{- 2}

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0.4 m s^{- 2}

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Solution

The correct option is D $0.4 m s^{- 2}$
From equation of motion,
$S = u t + \frac{1}{2} g t^{2}$
or, $5 = 0 \times 5 + \frac{1}{2} \times g \times (5)^{2}$
or, $5 = \frac{1}{2} \times g \times (5)^{2}$
or, $g = 0.4 m s^{- 2}$

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An object is dropped from a height of 5 m and it takes 5s to reach the ground. Calculate the acceleration due to gravity at that place.

The correct option is D 0.4ms−2From equation of motion, S=ut+12gt2 or, 5=0×5+12×g×(5)2 or, 5=12×g×(5)2 or, g=0.4ms−2

An object is dropped from a height of $5 m$ and it takes $5 s$ to reach the ground. Calculate the acceleration due to gravity at that place.

The correct option is D $0.4 m s^{- 2}$
From equation of motion,
$S = u t + \frac{1}{2} g t^{2}$
or, $5 = 0 \times 5 + \frac{1}{2} \times g \times (5)^{2}$
or, $5 = \frac{1}{2} \times g \times (5)^{2}$
or, $g = 0.4 m s^{- 2}$