Basketball throw physics Problem

  • Thread starter sundrops
  • Start date
  • #26
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YAY! It's right! Thanks SOOOO much vincent! you are a lifesaver - or at least a HUGE stress reliever! lol
 
  • #27
vincentchan said:
[tex]y = tan(38)x - \frac{g x^2}{2 v_{0}^2 cos^2(38)} [/tex]
How did you get this equation?
 
  • #28
Päällikkö
Homework Helper
519
11
ThetaInitial said:
How did you get this equation?
[tex]\left\{ \begin{array}{l}
y = y_0+v_{y0}t+\frac{1}{2}a_yt^2 \\
x = x_0+v_{x0}t+\frac{1}{2}a_xt^2 \\
\end{array} \right.
[/tex]
Solve for y:
[tex]\left\{ \begin{array}{l}
y = v_{y0}t+\frac{1}{2}a_yt^2 = v_0t \sin \theta -\frac{1}{2}gt^2 \\
x = v_{x0}t = v_0t \cos \theta \\
\end{array} \right.
[/tex]

[tex]t = \frac{x}{v_0 \cos \theta}[/tex]

[tex]y = \frac{v_0 \sin \theta x}{v_0 \cos \theta} -\frac{1}{2}g(\frac{x}{v_0 \cos \theta})^2[/tex]

[tex]y = x \tan \theta - \frac{gx^2}{2 v_0^2 \cos ^2 \theta}[/tex]
 
Last edited:

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