# Another Kinematics Problem

I'd really appreciate help on another kinematics problem. This one is harder than the first.

1. A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 70.0 m high (Fig. 2–34). (a) How much later does it reach the bottom of the cliff? (b) What is its speed just before hitting? (c) What total distance did it travel?

Okay. I understand everything conceptually. The stone is thrown up at a speed, and during that time it travels a certain distance above the cliff over a period of time before the stone reaches its highest point at which it's velocity is zero. Then, the rock falls the added distance and the 75 feet with earths acceleration of 9.8 meters/ s^2

I just don't know how to figure out what that added distance is. It is some 75+ x

Not sure what your question is exactly. 75 feet? Doesn't your problem say 70m?

BTW, have you learned energy yet? If so, try equating potential and kinetic energy to find some useful information. That will help you with some of the parts.

For (c), total distance, if you know the other parts, you should be able to figure out how high the stone reaches above the cliff. You may use kinematics or energy. If you know how high it went, you know that it traveled up and down that height, then the additional 70m to the cliff bottom.

djeitnstine
Gold Member
I'd really appreciate help on another kinematics problem. This one is harder than the first.

1. A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 70.0 m high (Fig. 2–34). (a) How much later does it reach the bottom of the cliff? (b) What is its speed just before hitting? (c) What total distance did it travel?

Okay. I understand everything conceptually. The stone is thrown up at a speed, and during that time it travels a certain distance above the cliff over a period of time before the stone reaches its highest point at which it's velocity is zero. Then, the rock falls the added distance and the 75 feet with earths acceleration of 9.8 meters/ s^2

I just don't know how to figure out what that added distance is. It is some 75+ x