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A hot-air balloon has just lifted off and is rising at the constant rate of 2.0 $\mathrm{m} / \mathrm{s} .$ Suddenly one of the passengers realizes she hasleft her camera on the ground. A friend picks it up and tosses itstraight upward with an initial speed of 13 $\mathrm{m} / \mathrm{s}$ . If the passengeris 2.5 $\mathrm{m}$ above her friend when the camera is tossed, how high isshe when the camera reaches her?

3.02 $\mathrm{m}$

Physics 101 Mechanics

Chapter 2

One-Dimensional Kinematics

Motion Along a Straight Line

Simon Fraser University

Hope College

University of Winnipeg

McMaster University

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Okay, so in this problem, we have a hot air balloon rising up. We have constant rates off too, meters per seconds on a person in the balloon and oppression in the ground. Tosses a camera. Two other person in the balloon with constant velocity. Initial velocity. Four of 13 you disburse. Second, the distance between the balloon and the ground was 2.5 meters, and the problem was to discover the final position off the balloon. That's mean. That's called about whom object A and the camera object to be. So the final position X is going to be the final position off the bottom. So physically, what we know is that when the camera reaches the balloon, the position of the balloon is equal the position of the camera. So we just need to describe the positions of the bathroom and the position off the camera and match both of them. Okay, so what is the position of the bottom about Lou moves with constant rate, which means it's going to be X. You're away plus VT and right is the positions of the camera. Ex me is just x zero b Poulos visi Vitti. Well, us U GT Square divided by two. We have to notice that the camera is under gravity. The separation which has miners nine point one as value. Okay, so we just need to match X A with X B. So let's do this. We're going to have X's youto, eh? This should be a a waas t because x zero b plus V zero T Poulos U t squared divided by two. Okay, what is the initial positions? Off the bottom? 2.5 velocity. Two initial positions off the camera. Zero because she wasn't a ground initial. Velocity. 13 miners, 9.81 divided by two T square. Okay, let's rearrange this. We're going to have minus nine point 81 t squared, plus 22 two minus five equals zero. This is a contract equation than we know how to solve what attracted equations. Let's do another page. We're just going to do the solution for the question. Active creation is going to be miners. 22 was minors. Dude square roots. 22 square miners Far times. Mine is 9.81 times minus five. All of these and the Squire roots. Apologies divided by two times minus nine points. 81. Okay, so let's see the solution Produce. We're going to have to solutions. Fresh solution is going to be time equals your 0.25 I've and the second solution. It's going to be 1.99. That point 99 seconds in seconds in here. Okay, So 1.99 this ray bigger. And we have to pick the first time. That is a solution. So we're going to solve the equation for a 0.2255 seconds. And the question that we have to solve is this one. We just want to discover the final position off the bottom. So let's do this. We have x A equals 2.5 plus two times judo, 0.25 fires. After multiplying all this, we have a final position off 3.2 uh, zero to meters. This is your final answer.

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