A stone is dropped from a cliff 490 m above its base. how long does the stone take to fall?
When a stone is dropped from a cliff, its initial velocity is equal to 0. The stone is accelerated in a direction vertically downwards at 9.8 m/s^2 due to the gravitational pull of the Earth.
Let the time taken by the stone to fall down a cliff 490 m high be t. Use the relation s = u*t + (1/2)*a*t^2 where s is the distance traveled, u is the initial velocity, a is the acceleration and t is the time taken.
For the stone that is dropped from the top of a cliff 490 m high, s = 490, a = 9.8 and u = 0.
Substituting these values gives:
490 = 0*t + (1/2)*9.8*t^2
=> t^2 = 490*2/9.8
=> t^2 = 100
=> t = 10 s
When the stone is dropped from a cliff 490 m high it takes 10 s to reach the base.