Calculating Initial Speed for a 45 Degree Angle Basketball Shot

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Homework Help Overview

The original poster is working on a physics problem involving projectile motion, specifically calculating the initial speed required for a basketball shot at a 45-degree angle. The scenario includes a player shooting from a height of 2 meters towards a basket that is 3.05 meters high, located 10 meters away.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants suggest writing equations for horizontal and vertical motion and emphasize the importance of using a common time variable. There are discussions about the quadratic nature of the equations involved and the potential for multiple solutions in projectile motion problems.

Discussion Status

Some participants have provided hints and guidance on how to approach the problem, including suggestions to simplify the height calculations. The discussion reflects a mix of attempts to derive equations and questions about the methods being used, with no clear consensus on the next steps.

Contextual Notes

There is mention of the original poster's struggle with quadratic equations and the complexity of the problem, including a reference to a degree 4 equation arising from their calculations. Participants are encouraged to share their working to facilitate further discussion.

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I got stuck in problem which states: "A basket ball player 2m high wants to make a goal from 10m from basket. If he shoots the ball at a 45 degree angle, at what initial speed must he throw the basketball. Height of basket is 3.05m." Please help.
 
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Sounds like homework so the mods might move this to the homework section. You should also use the homework template which has sections to fill in. Saves us time repeating stuff you already know.

Hint: Write equations for the horizontal and vertical motion. Then realize that the time at which the ball goes through the hoop is the same in both equations. Should give you enough equations and known variables to allow you to solve for the launch velocity. Perhaps start by subtracting 2m from all the heights so that the ball is launched from the ground towards a hoop at height 1.05m high.
 
CWatters said:
Sounds like homework so the mods might move this to the homework section. You should also use the homework template which has sections to fill in. Saves us time repeating stuff you already know.

Hint: Write equations for the horizontal and vertical motion. Then realize that the time at which the ball goes through the hoop is the same in both equations. Should give you enough equations and known variables to allow you to solve for the launch velocity. Perhaps start by subtracting 2m from all the heights so that the ball is launched from the ground towards a hoop at height 1.05m high.
I have tried but it gives quadratic equations which make solution even more complicated.
 
shayan haider said:
I have tried but it gives quadratic equations which make solution even more complicated.
More complicated than what? Than no solution at all?
Please post your working as far as you can get. Since, as CW pointed out, you have to use common time to connect the horizontal and vertical motions, solving a quadratic will be unavoidable.
 
Quite frequently there are two solutions to projectile problems. For example if you throw a ball there will be two occasions when it's at height X, once on the way up and once on the way down. So perhaps not surprising there is a quadratic to be solved sometimes.

Perhaps use..
http://www.purplemath.com/modules/quadform.htm
 
haruspex said:
More complicated than what? Than no solution at all?
Please post your working as far as you can get. Since, as CW pointed out, you have to use common time to connect the horizontal and vertical motions, solving a quadratic will be unavoidable.
I have derived equations for common time
CWatters said:
Quite frequently there are two solutions to projectile problems. For example if you throw a ball there will be two occasions when it's at height X, once on the way up and once on the way down. So perhaps not surprising there is a quadratic to be solved sometimes.

Perhaps use..
http://www.purplemath.com/modules/quadform.htm
I have an attempt in solving it. Please guide me how to carry on further.See the figure
 

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Can you show us your working to get to the equation in that image.
 
CWatters said:
Can you show us your working to get to the equation in that image.
What should I equate time to? I have equated it equal to " total time - time to reach 10m distance" but it gives equation of degree 4.
 
shayan haider said:
it gives equation of degree 4
Please post your working (sigh). We can't tell what you are doing wrong otherwise.
 

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