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?
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).