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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 65 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79

Problem 43 Hard Difficulty

SSM An airplane is flying with a velocity of 240 m/s at an angle of 30.0- with the horizontal, as the drawing shows. When the altitude of the plane is 2.4 km, a flare is released from the plane. The flare hits the target on the ground. What is the angle $\theta^{?}$

Answer

$\theta=41.50^{\circ}$

More Answers

06:59

Griffin G.

Related Courses

Physics 101 Mechanics

Physics

Chapter 3

Kinematics in Two Dimensions

Related Topics

Motion in 2d or 3d

Discussion

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DG

David Base G.

October 27, 2020

That was not easy, glad this was able to help

Alyssa J.

October 16, 2021

The wrong time is used! Instead, use t=37.54 because time is only allowed to be positive. The instructor in the video mistakenly used the wrong delta y, since Ay=-9.8m/s^2, you should use a -2400m for delta y because the flare is falling DOWN.

Alyssa J.

October 16, 2021

The correct answer is approximately 50.02 degrees

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Video Transcript

An airplane is flying with the velocity of 240 m/s at an angle of 36° with the horizontal. When the altitude of the plane is 2.4 km of flares released from the plane, the flare hits the target on the ground. What is the angle? Okay, So first we know that the angle is the inverse tangent of The Y Distance, which is 2400 m over the X distance. So since the acceleration in the X direction is equal to zero, that means X is equal to Vienna, co sign of 30° T. So we need to find T. First so we can find T. Because we know why is equal to negative abusive not Sign of 30° T plus one half a Y. T squared. So we know that wise negative 2400 m, We know a supply is negative 9.8 m/s squared. So this gives us 4.9 t squared plus 120 t -2400 equals zero. So using the quadratic equation, we get that the time is 13 seconds. Since the time is 13 seconds, the initial velocity is 240 m m per second. We get a distance of 2700 m. Since we have that distance of 2700 m, we get an angle of 42°.

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John D. Cutnell, Kenneth W. Johnson

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