Zero angle launch problem: Find height given inital speed and angle

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SUMMARY

The discussion centers on solving the zero angle launch problem for a basketball thrown horizontally at an initial speed of 4.40 m/s, landing at a point making a 30.0° angle with the horizontal. The relevant equations include the tangent function for angle calculation, the horizontal velocity equation (vfx = vix = 4.40 m/s), and the vertical motion equation (v2fy = v2iy - 2aΔy) with acceleration due to gravity set at -9.81 m/s². The solution approach involves deriving equations for horizontal and vertical motion, substituting y = xtanθ, and eliminating time (t) to find the release height.

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egadda2
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Homework Statement



A basketball is thrown horizontally with an initial speed of 4.40 m/s. A straight line is drawn from the release point to the landing point making an angle of 30.0o with the horizontal. WHAT WAS THE RELEASE POINT?

Homework Equations



tan[tex]\theta[/tex] = y/x
vfx= vix= 4.40 m/s
v2fy = v2iy - 2a[tex]\Delta[/tex]y

a = -g = -9.81
[c]3. The Attempt at a Solution [/b]
 
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Hi egadda2! :smile:

(have a theta: θ :wink:)

Find the equation for x and t, and the equation for y and t, then put y = xtanθ, and eliminate t. :wink:
 

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