Zero Launch Angle: Vox and Angle

[SteveThePharmer]

Homework Statement

A basketball is thrown horizontally with an initial speed of 4.20 m/s. A straight line drawn from the release point to the landing point makes an angle of 30.0° with the horizontal. What was the release height?

Vox=4.20 m/s=Vx Ax=0
Voy=0 m/s Ay=-9.81 m/s^2
Theta=180*-30.0* = 150*
Height=?

Homework Equations

tan [Theta] = (Vy/Vx)
Vy^2=Voy^2+2a[delta]y

The Attempt at a Solution

Tan[theta] x Vx = Vy
Tan 150* x 4.20 m/s = Vy = -2.42 m/s

Vy^2 = Voy^2 + 2a[delta]y; Voy=0
Vy^2 / 2a = [delta]y

(-2.42 m/s)^2 / -2(9.81m/s^2) = [delta]y = -0.298m

Which gives a height = 0.298m. So I must have done something wrong?

back of the book says answer should be 1.20m

 

[Doc Al]

tan [Theta] = (Vy/Vx)

That should be in terms of distances, not speeds: tan(theta) = Y0/x.

 

[ogb p]

Your calculations and approach seem to be correct. The only possible error could be in the conversion of degrees to radians. Make sure you are using radians instead of degrees in your calculations. If that is not the case, then it is possible that the answer in the book is incorrect. Double check your calculations and try again.