Question

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?

Answer 1

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

Answer 2

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.