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SteveThePharmer
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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
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