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
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

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.