What is the angle of a falling object's velocity after 2 seconds?

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A stone is thrown from a cliff at 30 m/s with a 0-degree angle, and the discussion focuses on determining its velocity angle after 2 seconds. The horizontal velocity (Vx) remains constant at 30 m/s, while the vertical velocity (Vy) is calculated using the equation Vy = Vyo + ay*t, resulting in a value of -19.6 m/s after accounting for gravity. The angle of the stone's velocity is derived from the tangent function, leading to the conclusion that the correct angle is approximately -33.2 degrees with respect to the horizontal. The discussion highlights the importance of using the inverse tangent function to find the angle accurately. The final calculations confirm that the angle is negative, indicating a downward trajectory.
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Homework Statement



A stone is thrown off a high cliff with speed 30m/s at an angle 0 degrees. At what angle is the stone's velocity 2 seconds after it's thrown. (measired with respect to the horizontal x direction(angle will be negative))

Homework Equations



angle = tan(Vy/Vx)



The Attempt at a Solution


V = velocity
Vx = stays constant = 30m/s
Vy = Vyo + ay*t = 30 m/s + 19.6 m/s = 49.6 m/s

angle = tan(49.6/30) which is wrong
 
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yankees26an said:
Vx = stays constant = 30m/s
Vy = Vyo + ay*t = 30 m/s + 19.6 m/s = 49.6 m/s
What's Vyo equal to?
 
Doc Al said:
What's Vyo equal to?

Then I guess it would be 0 m/s?

If it was 0 then I would have Vy = 19.6 and then

tan(19.6/30) is still not the right answer
 
Seems right to me. Don't forget that Vy and thus the angle are both negative.
 
tan((-19.6) / 30) = -0.765477483

The correct answer(supposedly) is -33.2 degrees with respect to the horizontal +x direction.
 
yankees26an said:
tan((-19.6) / 30) = -0.765477483
That should be inverse tan, not tan. tanθ = (-19.6/30) → θ = tan-1(-19.6/30). (I misread what you wrote before. :rolleyes:)
 

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