A stone is dropped on the moon from a height of 30 m- The acceleration due to gravity on the moon is g-6- Determine the approximate time taken by the stone to fall on the surface of the moon-Take g - 10 m-s2

Given : u = 0 m/s s = nbsp;30 m a = nbsp; g6 = 106 m/s^2 Using second equation of motion, s = ut + 12at^2 30 = 0 + ( 12×106× t^2 ) t^2 = 36 nbsp; t = 6 ...

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Standard IX

Physics

Displacement of COM:Application

A stone is dr...

Question

A stone is dropped on the moon from a height of $30 m$ . The acceleration due to gravity on the moon is $\frac{g}{6}$ . Determine the approximate time taken by the stone to fall on the surface of the moon.
(Take $g = 10 m / s^{2}$ )

2 s e c

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3 s e c

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5 s e c

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6 s e c

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Solution

The correct option is D $6 s e c$
Given :
$u = 0 m / s$
$s = 30 m$
$a = \frac{g}{6} = \frac{10}{6} m / s^{2}$

Using second equation of motion,
$s = u t + \frac{1}{2} a t^{2}$
$30 = 0 + (\frac{1}{2} \times \frac{10}{6} \times t^{2})$
$t^{2} = 36$
$t = 6 s e c$

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A stone is dropped on the moon from a height of 30 m. The acceleration due to gravity on the moon is g6. Determine the approximate time taken by the stone to fall on the surface of the moon. (Take g=10 m/s2 )

The correct option is D 6 secGiven : u=0 m/s s=30 m a=g6=106 m/s2 Using second equation of motion, s=ut+12at2 30=0+(12×106×t2) t2=36 t=6 sec

A stone is dropped on the moon from a height of $30 m$ . The acceleration due to gravity on the moon is $\frac{g}{6}$ . Determine the approximate time taken by the stone to fall on the surface of the moon.
(Take $g = 10 m / s^{2}$ )

The correct option is D $6 s e c$
Given :
$u = 0 m / s$
$s = 30 m$
$a = \frac{g}{6} = \frac{10}{6} m / s^{2}$

Using second equation of motion,
$s = u t + \frac{1}{2} a t^{2}$
$30 = 0 + (\frac{1}{2} \times \frac{10}{6} \times t^{2})$
$t^{2} = 36$
$t = 6 s e c$