If Mere an astronaut on the moon drops a hammer from a height of 1.4 meters. How long will it take the hammer to reach the ground if the gravitational acceleration on the moon is (one sixth) of that on earth?

2 Answers
J W.
Feb 25, 2018

1.30 seconds.

Explanation:

Assuming the acceleration due to gravity on Earth is an exact value #-9.8 m/s^2#, one sixth of this value is given by #(-9.8m/s^2) * 1/6 = -1.63 m/s^2#

This value is our new a.

But #a = -g.#

Using formula, where h is height, t is time taken, and g is the magnitude of the acceleration,

#h = 1/2g(t)^2 #

and solving for time with algebraic manipulation we get:

#t= sqrt(2h)/(g) #

Plug in our givens and get

#t= sqrt(2(1.4m))/(1.63m/s^2) #= 1.30 seconds

Athul
Feb 25, 2018

1.3 s

Explanation:

Let acc. on the moon be g'
As acceleration due to gravity on the moon is #(1/6)^(th)# of that on the earth,
g'= g/6. ( where g is the acc. due to gravity on the earth)
Now, according to the second law of kinematics,
S=#ut+1/2at^2#
Here, s= #1.4# m, a=g', u=0m/s (free fall)
So we have, #1.4#= #0*t+ 1/2 g' t^2#
#1.4 =1/2 g/6 t^2#
As g= 9.8 #ms^-2#
On simplifying we get, #1.714 = t^2#
#implies t =sqrt(1.714)# = #1.3# s.

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