Physics? gravity and velocity?

Question

a stone is thrown vertically downward from a 200 m high cliff at an initial velocity of 5 m/s.

A. How long does it take for the stone to reach the base of the cliff? B. What is the stone's final velocity?

please explain. thank you

Answer 1

vo= initial velocity  (5 m/s)

g = acceleration    (9.8 m/s^2)

Distance:  d = vo+ 0.5 * g * t^2

200 = 5+0.5 *9.8*t^2

200 = 9.9*t^2

t^2 = 200/9.9

t^2 = 20.202

t = SQR 20.202

t = 4.49s  (time to reach the base of the cliff)

Velocity:  v = vo+ g * t

v = 5+(9.8*4.49)

v = 5+44

v = 49m/s  (stone's final velocity)

Answer 2

IT takes 40seconds and the stone's final velocity is still 5m/s as falling things will remain at a constant speed unless more force is added to the ongoing direction.

Plus, i tried to make this as less chimp as possible. You should see the proper principle, it is so chimp that you couldn't even understand it!

Answer 3

using

v^2=u^2+2as

where v is the final velocity and u the initial velocity

v^2=25 + 2*9.8*200

v^2=3945

v=62.8 m/s

using

v=u+at

62.8=5+9.8t

t=5.9 secs

ignoring air resistance