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. see the chart below and me if you can. you .

consider the motion in y-direction

v₀ = initial velocity = 0 m/s

a = acceleration = g = - 9.8 m/s²

t = time

v = final velocity at any time "t"

velocity at any time is given as

v = v₀ + at

v = 0 + (- 9.8) t

v = (- 9.8) t

so at t = 1  , v = (- 9.8) (1) = - 9.8 m/s

at t = 2  , v = (- 9.8) (2) = - 19.6 m/s

at t = 3  , v = (- 9.8) (3) = - 29.4 m/s     and so on

Y₀ = initial position where the marble was dropped from = 0 m

Y = final position at any time "t"

final position at any time "t" is given as

Y = Y₀ + v₀ t + (0.5) a t²

Y = 0 + (0) t + (0.5) (-9.8) t²

Y = - 4.9 t²

at t = 1 , Y = - 4.9 (1)² = - 4.9 m

at t = 2 , Y = - 4.9 (2)² = - 19.6 m

at t = 3 , Y = - 4.9 (3)² = - 44.1 m  

and So on

the skeletons of most mammals reach a certain size and then stop growing. however, many animals, including some mammals, keep growing throughout their lives. kangaroos, for example, just keep growing and growing until they die. also, the longer an animal lives the greater the chance it might catch some deadly disease.