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\cdots An airplane is flying with a velocity of 90.0 $\mathrm{m} / \mathrm{s}$ at an angle of $23.0^{\circ}$ above the horizontal. When the plane is 114 $\mathrm{m}$ directly above a dog that is standing on level ground, a suitcase drops out of the lugage compartment. How far from the dog will the suitcase land? You can ignore air resistance.

795 m

Physics 101 Mechanics

Chapter 3

Motion in a Plane

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Kent L.

October 28, 2020

University of Washington

University of Winnipeg

McMaster University

Lectures

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we have the initial velocity of the aircraft 90 meters per second. We have a angle of 23 degrees above the horizontal and me know that delta Y equals 114 meter. So ah, suitcase was dropped out of the aircraft at a 114 meters and directly below is a dog. So they're asking us thie distance between the suitcase once it hits the ground and the dog assuming that the dog and the suitcase are both on the ground, the only distance the only displacement here would be in the ex direction between the two objects. So we can say that Delta y equals V Y initial times t plus 1/2 of g t squared. We're choosing upwards to be positive. So we'll say negative. Rather, 114 equals 90 sign of 23 degrees times T minus 4.9 t squared. So here gravity is going to be negative and this is a quadratic formula, so you can plug this into your tia 80 40 85 2 89 in order to solve for tea, and we find that tea is going to be equal to 9.60 seconds. This would be the time and on the air for the luggage. So we can say that Delta X is going to be equal to the ex initial T plus 1/2 A T squared in the ex direction. There isn't any acceleration. So this term become zero and we find that 90 co sign of 23 degrees times 9.60 seconds. This will equal your total horizontal displacement. This will equal approximately 795 meters. So we can say that Theseus case is 77 195 meters from the dog. That is the end of the solution. Thank you for watching.

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