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

[yankees26an]

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

 

[Doc Al]

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?

 

[yankees26an]

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]

Seems right to me. Don't forget that Vy and thus the angle are both negative.

 

[yankees26an]

tan((-19.6) / 30) = -0.765477483

The correct answer(supposedly) is -33.2 degrees with respect to the horizontal +x direction.

 

[Doc Al]

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:)