Question

A stone is dropped from the edge of a cliff. What is the velocity of the stone after 2.5 seconds and how far does it fall?

Answer 1

Answer:

Explanation:

The acceleration due to gravity is ~ 9.8 m/ss, let’s just use 10 for simplicity.

We can use the kinematic equations to find the answers to your question.

However, we are assuming here that the rock is falling in a vacuum.

Firstly the velocity 3 seconds later we can use:

V(f) = V(I) + at

f for final, i for initial, a acceleration, t time

V(f) = 0 + 10 * 3

= 30 m/s

Then for the second part we can use this equation to find the distance traveled in the 4th second only:

S = V(i) * t + (1/2) * a * t(^2)

Where s is the distance traveled

S = 30 * 1 + 1/2 * 10 * 1

= 35 meters!

Remember this is the disctace in the 4th second, not the total distance over all time. (80 meters is the total if you were wondering)