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(III) A ball is dropped from the top of a 50.0 -m-high cliff. Atthe same time, a carefully aimed stone is thrown straight upfrom the bottom of the cliff with a speed of 24.0 $\mathrm{m} / \mathrm{s}$ . Thestone and ball collide part way up. How far above the baseof the cliff does this happen?

$y_{f}=28.7$ meters above the base of the cliff when the ball and stone collide

Physics 101 Mechanics

Chapter 2

Describing Motion: Kinematics in One Dimension

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

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03:28

Newton's Laws of Motion are three physical laws that, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. These three laws have been expressed in several ways, over nearly three centuries, and can be summarised as follows: In his 1687 "Philosophiæ Naturalis Principia Mathematica" ("Mathematical Principles of Natural Philosophy"), Isaac Newton set out three laws of motion. The first law defines the force F, the second law defines the mass m, and the third law defines the acceleration a. The first law states that if the net force acting upon a body is zero, its velocity will not change; the second law states that the acceleration of a body is proportional to the net force acting upon it, and the third law states that for every action there is an equal and opposite reaction.

04:16

In mathematics, a proof is a sequence of statements given to explain how a conclusion is derived from premises known or assumed to be true. The proof attempts to demonstrate that the conclusion is a logical consequence of the premises, and is one of the most important goals of mathematics.

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I'm gonna draw a diagram for this problem just to make it a little clear what I'm doing. So we have a ball that's at the top of the cliff, which is 50 meters high, and a stone is thrown upwards as the ball is dropped, its thrown upwards of 24 meters per second. And at some point they collide. And I'm gonna call this point, why final? And now I'm just gonna list everything I know for the stone and the ball. The stone is starting at the ground. The ball is starting at the top of the cliff. So initial height of the stone is zero and it shall have the ball. It's 50 meters. We also know why finals the same for both. We know our initial velocity for the stone is 24 meters per second and our initial velocity for the ball. It's starting restaurants dropped, so it's Notre Velocity zero. We also know that a equals minus G. That's true for both of them, cause they're in free fall. And now I'm gonna write an equation for both of them. I'm gonna read it in terms of y f and not Delta y So why F equals, uh, y zero plus V zero t plus 1/2 A t squared. And if I were to just bring this y zero to the other side, we would get adults. Why? So it's it's the same equation. Now for the stone we know the initial height is zero. The initial velocity is 24 and this is a minus g. She squared. Eso y final. This is the expression for the stone is gonna be 24 t minus 1/2 geechee squared and for the ball, we're starting at 50 meters. We have no velocity. And again minus 1/2 Cici squared. So for the ball, our expression is this. And now we can just set these two equal to each other and sell for time. So 24 t and you can see now that these two terms, we're gonna cancel nicely. So there is gone and we get 24 t equals 50. So T is around 2.83 But I'm not going around yet. Uh, and now we have the time at which they collide. We want to know the height so we can plug into either equation we got, But I'm just gonna use this one because there's only one t in there I don't know looking and twice. So our wife final, it's gonna be 50 minus 1/2. She is 9.8 and our t 2.83 squared and our wife final ends up being 28.7 meters and this is 28.7 meters above the ground. It didn't matter which equation I plugged into. If I did the other one, I would still get 28.7 meters. Either way, they collide at the same point and this is tight.

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