Calculating the Distance and Velocity of a Thrown Object Using Kinematics

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A stone is thrown vertically upward at 12.0 m/s from a 70.0 m high cliff, prompting questions about its time to reach the bottom, speed before impact, and total distance traveled. The stone first ascends until its velocity reaches zero, then descends back down, adding to the total distance. To find the maximum height above the cliff, one can use kinematics or energy principles. The total distance traveled is the sum of the height reached above the cliff and the 70.0 m drop. Clarification on the reference to "75 feet" is needed, as the problem specifies 70 m.
crimsonn
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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 Earth's 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

help please
 
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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.
 
crimsonn said:
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 Earth's 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

help please

What is the maximum height when V = 0 ? Then add that to the height of the cliff.
 
It would be x + 70.0m. Not sure where "75 feet" is coming from.
 

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