A 4.6 kg rock is dropped from rest down a vertical mine shaft.  How many meters will the rock have dropped after 4.24 s?

Expert Answers

An illustration of the letter 'A' in a speech bubbles

Hello!

I suppose we ignore air resistance. "Dropped from rest" means zero initial speed. Then the depth of the fall is given by the formula

`H(t) = (g*t^2)/2,`

where t is for time and g is the gravity acceleration, g = approx. 9.8 `m/s^2` .

So after 4.24s the depth...

Unlock
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Start your 48-Hour Free Trial

Hello!

I suppose we ignore air resistance. "Dropped from rest" means zero initial speed. Then the depth of the fall is given by the formula

`H(t) = (g*t^2)/2,`

where t is for time and g is the gravity acceleration, g = approx. 9.8 `m/s^2` .

So after 4.24s the depth of the rock will be

`(9.8*4.24^2)/2` = 88.09 (m).

Approved by eNotes Editorial Team