(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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=?

2. Relevant equations

tan [Theta] = (Vy/Vx)

Vy^2=Voy^2+2a[delta]y

3. 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

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# Homework Help: Zero Launch Angle: Vox and Angle

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