Question

A stone is dropped from rest from the top of a building. It takes Δt = 2.2 s for it to reach the ground.

Part (a) What is the initial velocity, vi, of the stone in m/s?
Part (b) What is the value of the magnitude of acceleration, in m/s^2?

Answer 1

Value of magnitude of acceleration will be
`9.8m/sec^2`

Step-by-step explanation:

It is given that when a stone is dropped it takes 2.2 sec to reach the ground

(a) As the stone is dropped from the top of building

So its initial velocity
`u_i` will be 0 m /sec

(b) As the stone is free falling and there is no external force applied on it so its acceleration will be equal to acceleration due to gravity

So value of magnitude of acceleration will be equal to
`9.8m/sec^2`