A feather is dropped on the moon from a height of 10 m. The acceleration due to gravity on the moon is $1.67 m s^{- 2}$ . Determine the time taken by the feather to fall on the surface of the moon.

3 s

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4.12 s

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3.46 s

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1.41 s

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Solution

The correct option is C $3.46 s$
Given, distance covered (height) $= 10 m$ , acceleration $a = 1.67 m s^{- 2}$ and initial velocity, $u = 0$
Let the time taken be 't'.
Using the second equation of motion,
$s = u t + \frac{1}{2} a t^{2}$
$10 = 0 + \frac{1}{2} \times 1.67 \times t^{2}$
$\Rightarrow t = \sqrt{\frac{20}{1.67}}$ s
$\Rightarrow t = 3.46 s$

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A feather is dropped on the moon from a height of 10 m. The acceleration due to gravity on the moon is 1.67 ms−2​. Determine the time taken by the feather to fall on the surface of the moon.

The correct option is C 3.46 sGiven, distance covered (height) =10 m, acceleration a=1.67 ms−2 and initial velocity, u=0 Let the time taken be 't'. Using the second equation of motion, s=ut+12at2 10=0+12×1.67×t2 ⇒t=√201.67 s ⇒t=3.46 s

A feather is dropped on the moon from a height of 10 m. The acceleration due to gravity on the moon is $1.67 m s^{- 2}$ . Determine the time taken by the feather to fall on the surface of the moon.