physics equation.... please help (1 Viewer)

[The_Chosen_One]

If I dropped a 40kg rock over the edge of a 60m high cliff on the Moon, where g=1.64ms-2

a) How long would the rock take to hit the ground and
b) What would the rock's weight be on the Moon?

 

[youngminii]

I'm assuming the 40kg is when it's measured on Earth.
Weight of rock on Earth = 40kg
Mass of rock on Earth = 40/9.8 = 4.08163..
Weight of rock on Moon = 4.09163.. x 1.64 = 6.693877.. = 6.7kg (2 sf)

As for a), lol I can't remember the formula for one dimonsional projectile motion. Just sub it in.

 

[Drongoski]

If I dropped a 40kg rock over the edge of a 60m high cliff on the Moon, where g=1.64ms-2

a) How long would the rock take to hit the ground and
b) What would the rock's weight be on the Moon?

Are the answers:

a) 8.554 s

b) 65.6 N

??

 

[lolokay]

Are the answers:

a) 8.554 s

b) 65.6 N

??

this is what i get too

 

[Drongoski]

this is what i get too

Since you are pretty smart, I guess my answer is prob correct. Good to remember a bit of mechanics after all these years.

 

[Drongoski]

I'm assuming the 40kg is when it's measured on Earth.
Weight of rock on Earth = 40kg
Mass of rock on Earth = 40/9.8 = 4.08163..
Weight of rock on Moon = 4.09163.. x 1.64 = 6.693877.. = 6.7kg (2 sf)

As for a), lol I can't remember the formula for one dimonsional projectile motion. Just sub it in.

40 kg is mass; mass is independent of location.

Weight is the force on the mass due to gravity; the g on earth is approx 9.8 m/s/s (is that right ?) whereas on the the moon it's given as 1.64

Therefore weight on earth = F = ma = 40 x 9.8 Newtons
weight on moon = F = ma = 40 x 1.64 Newtons

As for working out the time 't', you can use:

v^2 = u^2 + 2as = 0^2 + 2 x 1.64 x 60 = 196.8

v = u + at => sqrt(196.8) = 0 + 1.64 x t etc