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(III) A toy rocket moving vertically upward passes by a2.0 -m-high window whose sill is 8.0 $\mathrm{m}$ above the ground. Therocket takes 0.15 $\mathrm{s}$ to travel the 2.0 $\mathrm{m}$ height of the window.What was the launch speed of the rocket, and how high will itgo? Assume the propellant is burned very quickly at blastoff.

$v_{o}=18.8$ meters per seconds$y_{\text {total}}=18.1$ meters

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

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

University of Sheffield

Lectures

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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Okay, so here, we're gonna say upwards is positive on here. We're trying to find, uh, the velocity at the bottom of the window for the toy and then the velocity at lunch. So we can say that if upwards is positive, the acceleration would simply be equal to G, which would be equal to negative 9.0 meters per second squared. And if this is the case, we can choose the ground to be the wine. The Y final equals zero meters. That would be the ground. Um and we need to just get an intermediate result. Inarritu's for solved for the velocity at launch. So we can say that why the position of top of window would be equal to the Y position at the bottom of the window, plus the velocity at the bottom of the window multiplied by the time it takes to pass the window and then plus 1/2 g times t the time it takes to pass the window quantity squared, and so we can solve. So we can say that 10 meters equals 8.0 meters plus velocity at the bottom of the window. Times 0.15 seconds plus rather say rather weaken Sue minus 4.9 times 0.15 seconds. Quantity squared. And so we can simply use the saw function on your T 84 85 or 89 in order to solve for the velocity at the bottom of the window. And this will equal 14 0.7 meters per second s. So this is just initial media result. We we needed to find this in order to find the velocity of launch. So now we can use the velocity at the bottom of the window in order to find the launch velocity. Assuming that the launch philosophy was of course, you know, uh, at zero meters so we can say that velocity at the bottom of the window squared which would again equal, be final squared. Essentially, we could be at launch. So the initial squared plus two g times also why Or in this case, it will be why bottom of and we can see that velocity at launch gonna be equal to the square root of the velocity of the bottom of window quantity, squared minus or rather, weaken, simply say, plus to g times. Why at the bottom of the window. Rather my apologies. It will be minus my apologies. It's realizing that, uh, closed for the seats all to the 1/2 power. And at this point now we can solve so velocity at launch would be equal to the velocity we just found. So 14.7 meters per second quantity squared AA plus two times 9.80 meters per second squared times 8.0 meters all to the 1/2 power. We find that the velocity at launch is going to be equal to approximately 18.8 meters per second. This is our final answer. That is the end of Oh, my apologies, we This is the velocity launch. We need to find the maximum height. So last part. To find the maximum height, we find the velocity at max Height, quantity squared. This is gonna equal velocity at launch squared plus two times a times. Why, Max or rather, G. Why Max minus y initial. Ah, we're going to say that we know the velocity at the maximum height is gonna be zero because it is always zero. So we can say that why Max would be equal to y initial plus the velocity at Rather, we can say plus negative velocity at lunch. Quantity squared, divided by two times G. This has also been equal zero because we're starting from the ground and we've Are we already in the last part of the equation were actually switching our perspective that the why initial would be at the ground level so this would be equal to serial meters and this would be equal to negative 18.84 meters per second quantity squared, divided by two times negative of 9.80 meters per second squared and we find that the why Max would be 18.1 meters, so this would be the maximum height reached. That is the end of the solution. Thank you for watching.

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