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I give up! Basketball Projectile problem.

by kieth89
Tags: basketball, projectile
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kieth89
#1
Sep9-12, 12:10 AM
P: 23
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
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SammyS
#2
Sep9-12, 12:51 AM
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Quote Quote by kieth89 View Post
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
What are the conditions for which this range formula is valid?
[itex]\displaystyle
R = \frac{2v_{0}^{2}\sin \theta \cos \theta}{g}\ ?[/itex]
It'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.
kieth89
#3
Sep9-12, 01:34 AM
P: 23
Quote Quote by SammyS View Post
What are the conditions for which this range formula is valid?
[itex]\displaystyle
R = \frac{2v_{0}^{2}\sin \theta \cos \theta}{g}\ ?[/itex]
It'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.
Plugging the numbers into that formula results in taking the square root of a negative..and there should be no i in this Physics problem. With my values plugged in the equation looks like this (for the high end):

[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...

SammyS
#4
Sep9-12, 01:56 AM
Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,801
I give up! Basketball Projectile problem.

Quote Quote by kieth89 View Post
Plugging the numbers into that formula results in taking the square root of a negative..and there should be no i in this Physics problem. With my values plugged in the equation looks like this (for the high end):

[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...
Well, you're right! That final result is wrong (in the link). Their initial equations for x & y look good.

[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 v0.
kieth89
#5
Sep9-12, 03:10 AM
P: 23
Quote Quote by SammyS View Post
Well, you're right! That final result is wrong (in the link). Their initial equations for x & y look good.

[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 v0.
Thanks for the help..I figured it out, but the program didn't like how I formatted my answer (gave it the range instead of the difference between the two extremes) so I still got it wrong :(...At least I know how to do it for the test now though.


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