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Question: A feather is dropped on the moon from a height of 1.40 meters. The acceleration due to gravity on th...

A feather is dropped on the moon from a height of 1.40 meters. The acceleration due to gravity on the moon is 1.67 ms2m{{s}^{-2}}. Determine the time for the feather to fall to the surface of the moon. (in sec.)
A. 1.29 s
B. 1.2 s
C. 1.9 s
D. 0.29 s

Explanation

Solution

There is no “air resistance” on the moon’s surface. Thus any object that may be a feather, or a stone, irrespective of the weight of the body, the formula of linear motion (the distance - acceleration) is applicable.
Formula used:
s=ut+12at2s=ut+\dfrac{1}{2}a{{t}^{2}}

Complete step-by-step solution:
From given, we have the data,
The height from which a feather is dropped on the moon, h = 1.4 m.
The acceleration due to gravity on the moon, a = 1.67 ms2m{{s}^{-2}}.
The initial motion of the feather, u = 0 m/s
To determine: The time for the feather to fall to the surface of the moon, t =?
The derived Newton’s second law of motion formula is,
s=ut+12at2s=ut+\dfrac{1}{2}a{{t}^{2}}
Where, uuis the initial velocity of the object/body, aais the acceleration of the object/body, ttis time taken and ssis the displacement.
In this given problem, consider the height as the displacement.
So, s = h = 1.4 m.
Substitute the given values in the formula of Newton’s second law.
1.4=(0)t+12(1.67)(t2)-1.4=(0)t+\dfrac{1}{2}(-1.67)({{t}^{2}})
The negative signs represent the direction.

& -1.4=-0.835{{t}^{2}} \\\ & {{t}^{2}}=\dfrac{1.4}{0.835} \\\ & t=1.29\,s \\\ \end{aligned}$$ Thus, the value of time taken is 1.29 s. **As the value of time obtained is 1.29 s, thus, the time for the feather to fall to the surface of the moon is 1.29 s. Option (A) 1.29 s is correct.** **Note:** The things to be on your figure tips for further information on solving these types of problems are: The units of the given parameters should be checked first before proceeding to solve the problems. If, for example, in this question, the height and the acceleration were given in terms of meters and centimeters, then should change the unit of anyone, to get the same unit.