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I give up! Basketball Projectile problem. 
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#1
Sep912, 12:10 AM

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1. The problem statement, all variables and given/known data
A basketball leaves a player's hands at a height of 2.20 above the floor. The basket is 2.90 above the floor. The player likes to shoot the ball at a 40.0 degree angle. If the shot is made from a horizontal distance of 6.00 and must be accurate to plus or minus .22 m(horizontally), what is the range of initial speeds allowed to make the basket? Express your answer using two significant figures. 2. Relevant equations 2d kinematic equations and projectile equations. 3. The attempt at a solution I'm really tired and sick of this problem, and only have one more try I can use in the homework problem. So far I've tried: following the basic method used here: http://answers.yahoo.com/question/in...0112245AAlbiKZ , but it didn't work.. So then I tried here: http://answers.yahoo.com/question/in...2202754AAuIjr2 , but it didn't work either. So then I used my own derived formula for range : [itex] R = \frac{2v_{0}^{2}sin \Theta cos \Theta}{g}[/itex]...and using this didn't work either. And now I only have one more try before I'm done with the problem. So I need to get it right. Honestly, I thought my Physics class would be kind of fun and interesting, but right now I absolutely hate it and have no idea how to solve these problems. I'm going to get help at my college Monday, but this homework is due tomorrow, so I need to figure this problem out. Thanks for any help, Josh 


#2
Sep912, 12:51 AM

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[itex]\displaystyleIt's valid for the case in which the final vertical position is the same as the initial vertical position. That's not the case here. What is the horizontal equation of motion appropriate for this problem ? What is the verticalal equation of motion appropriate for this problem ? Added in Edit: The first link you gave gives the solution to essentially the same problem, with a couple of the numbers changed. What was you problem with working with that method? It was done correctly. 


#3
Sep912, 01:34 AM

P: 23

[itex]v_{0} = \sqrt{6.22 tan(40°)  9.8 x^{2} (\frac{1 + tan(40°)^{2}}{2*0.7}}[/itex] Which ends up imaginary... 


#4
Sep912, 01:56 AM

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I give up! Basketball Projectile problem.
[itex]\displaystyle \Delta x=v_0\cos(\theta)t[/itex] [itex]\displaystyle \Delta y=v_0\sin(\theta)t\frac{1}{2}gt^2[/itex] Solve the first equation for t. Plug that result into the second equation, then solve for v_{0}. 


#5
Sep912, 03:10 AM

P: 23




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