# Angle of falling object

1. Dec 18, 2009

### yankees26an

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

2. Dec 18, 2009

### Staff: Mentor

What's Vyo equal to?

3. Dec 18, 2009

### yankees26an

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

4. Dec 18, 2009

### Staff: Mentor

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

5. Dec 18, 2009

### yankees26an

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

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

6. Dec 18, 2009

### Staff: Mentor

That should be inverse tan, not tan. tanθ = (-19.6/30) → θ = tan-1(-19.6/30). (I misread what you wrote before. :uhh:)