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Projectile motion of basketball

  • Thread starter tensor0910
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  • #1
tensor0910
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Homework Statement


trying to figure out how to calculate the angle of release needed to shoot a basketball in a rim. Here's the data I have so far:

distance from shooter to rim: 6.28m
height of rim: 3m
height of shooter: 1.75m
Initial velocity at time of release: 8.68 m/s ( hypotenuse of triangle )
height ball is released from shooter: 2.13m
height from release of ball till level plane with rim: .86m


Homework Equations


[/B]
Vff = V2i + 2aΔx
Vf = Vi + at
ax = 0
ay = 9.8


The Attempt at a Solution


[/B]
The shooter is much shorter than the typical projectile motion parabola so we have to figure out at what distance from the rim does the parabola actually start. I used a motion tracking program to figure out the velocity of the release of the ball. But its released so I only have the hypotenuse. This is where I am stuck.

Any and all comments are appreciated
 

Answers and Replies

  • #2
Simon Bridge
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The shooter is much shorter than the typical projectile motion parabola so we have to figure out at what distance from the rim does the parabola actually start.
... what do you mean: you only need to know 3 points on the parabola to determine it.
Besides - why not "start" the parabola at the position of the ball when the shooter releases it?

I used a motion tracking program to figure out the velocity of the release of the ball. But its released so I only have the hypotenuse. This is where I am stuck.
You have the magnitude of the initial velocity, but not it's direction? Is this correct?
Isn't the direction what you want to calculate?
 
  • #3
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I think restating the problem might help, for an angle ##\theta## at a time ##t## you need the vertical displacement to be 0.86 m and horizontal displacement to be 1.75 m. Use the three kinematic equations to find ##\theta## and you're done.
 
  • #4
haruspex
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  • #5
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6.28m?
Yeah right, I misread it. Thanks.
 
  • #6
epenguin
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I don't see the diameter of the ball mentioned - don't you think this is essential in modelling this system? I.e. where exactly does the ball have to arrive?

Do you expect a unique solution?

You seem to be expected to model this as frictionless - how good do you expect that to be?
 
  • #7
tensor0910
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... what do you mean: you only need to know 3 points on the parabola to determine it.
Besides - why not "start" the parabola at the position of the ball when the shooter releases it?

You have the magnitude of the initial velocity, but not it's direction? Is this correct?
Isn't the direction what you want to calculate?
I was working under the impresssion that the parabola needed to have a flat bottom in order for us to make our calculations.

Im trying to calculate the angle of release needed for the ball to make it in the basket. With the distances given we can assume the ball went in.

Simon Bridge you are correct I have the initial velocity but im not sure how to find the angle since the ball goes in the rim at a higher distance than the shooter.
 
  • #8
tensor0910
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I don't see the diameter of the ball mentioned - don't you think this is essential in modelling this system? I.e. where exactly does the ball have to arrive?

Do you expect a unique solution?

You seem to be expected to model this as frictionless - how good do you expect that to be?
epenguin thank you for taking the time to respond; I didnt mention the diameter of the ball. I should have stated that with the distances listed it is the ball went in the rim- my mistake poor wording on my part. Friction will be negligible in this model.
I expect the answer to be somewhere between 45-55 degrees because thats the avg. release point for a basketball. Finding it is just the trouble.
 
  • #9
gneill
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You need to write and solve the projectile motion equations for the given situation. You can simplify things a bit by just considering the actual Δx and Δy that needs to take place between the launch instant and the instant the ball reaches the target location.
 
  • #10
Simon Bridge
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I was working under the impresssion that the parabola needed to have a flat bottom in order for us to make our calculations.
Whatever gives you that idea?

I think there may be a conceptual issue here:
The parabola in question has a maxima, but no minimums - nothing that can be called a "base" ... so how can it have a "bottom".

Are you giving special status to the ground here? (What if the basketball court was on the roof of a skyscraper: would you need the height of the skyscraper too?) Why not put "the bottom" at the height that the ball was released from? Why do you need the whole parabola at all, you only need the start and the end and some idea of the relationship between them right?

If you need to visualise the motion - draw a horizontal line for the ground, draw a point for where the ball starts it's journey, and another where the ball passes through the hoop (this will be higher than the 1st point and off to one side) ... sketch a parabola through both points... notice that you can sketch many different parabolas through both points. Those are all valid solutions, so you will need to use the information you have to narrow down the possibilities.

Im trying to calculate the angle of release needed for the ball to make it in the basket. With the distances given we can assume the ball went in.
Simon Bridge you are correct I have the initial velocity but im not sure how to find the angle since the ball goes in the rim at a higher distance than the shooter
I think I see your confusion - you have a set of worked out ballistic equations that assume that the ball ends it's flight at the same height it is released and you want to use those to help solve the problem.
Best practice is to not do that ... solve the kinematic equations for this specific problem instead ... it can help to sketch v-t diagrams for the x and y components of the motion.
 
  • #11
tensor0910
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Simon Bridge you hit it right on the nose. I need to break this down into two different equations. Thank you for the help everyone!
 

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