Amarble is dropped off the top of a building. find it's velocity and how far below to its dropping point for the first 10 seconds of the drop .use g=-9.8m/s^2

1) velocity: 98 m/s downward

Explanation:

The marble moves by uniformly accelerated motion, with constant acceleration a=g=-9.8 m/s^2 directed towards the ground. Therefore, its velocity at time t is given by:

where

is the initial velocity

is the acceleration

is the time

Substituting t = 10 s, we find:

And the negative sign means the direction of the velocity is downward.

2) Distance covered: 490 m

The distance covered in an uniformly accelerated motion can be found with the formula:

where

is the acceleration

t is the time

Substituting t=10 s, we find

And the negative sign means the displacement is below the dropping point.

i already answered it by 34/s