Angle of falling object

1. The problem statement, all variables and given/known data

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))

2. Relevant equations

angle = tan(Vy/Vx)



3. 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
 
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
 

Doc Al

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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.
 

Doc Al

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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. :uhh:)
 

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