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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 $\theta^{?}$

$\theta=41.50^{\circ}$

06:59

Griffin G.

Physics 101 Mechanics

Chapter 3

Kinematics in Two Dimensions

Motion in 2d or 3d

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.

The correct answer is approximately 50.02 degrees

Hope College

University of Sheffield

University of Winnipeg

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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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