Basketball Projectile Motion Problem

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SUMMARY

The basketball projectile motion problem involves calculating the initial speed required for a player, standing 10.0 meters away from a basket 3.05 meters high, to successfully shoot the ball at a 40.0-degree angle without hitting the backboard. The relevant equations used are the kinematic equations for horizontal and vertical motion: xf = xi + vit + 1/2at² and yf = yi + vit + 1/2at². The resulting quadratic equation derived from these equations is -4.9 (10/vi(sin40))² + (vi(cos40))(10/vi(sin40)) - 1.05 = 0, which needs to be solved for the initial speed vi. The discussion also touches on formatting equations using LaTeX for clarity.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Basic knowledge of projectile motion concepts
  • Familiarity with trigonometric functions (sine and cosine)
  • Experience with solving quadratic equations
NEXT STEPS
  • Learn how to solve quadratic equations using the quadratic formula
  • Study the principles of projectile motion in physics
  • Explore LaTeX formatting for mathematical equations
  • Investigate the effects of angle and initial speed on projectile trajectories
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Students studying physics, educators teaching projectile motion, and anyone interested in applying mathematical concepts to real-world scenarios such as sports physics.

Jeff231
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Homework Statement



A basketball player who is 2.00[m] tall is standing on the floor 10.0 [m] from the basket, as in the figure. If he shoots the ball at 40.0degrees angle with the horizontal, at what initial speed must he throw so that it goes through the hoob without striking the backboard? The basket height is 3.05[m].

physicsc.jpg



Homework Equations



xf = xi + vit + 1/2at^2
yf= yi +vit + 1/2at^2

The Attempt at a Solution



I used the yf equation to come up with an equation that has vi and t as the variables. Then I used the xf equation and solved for t. Then I plugged in that t into the yf equation. It comes out to a pretty large quadratic equation:

-4.9 (10/vi(sin40))^2 + (vi(cos40))(10/vi(sin40)) - 1.05 = 0

Now I'm having trouble finding vi. I'm pretty sure I did everything right, but can someone help me figure out how to solve for vi?

Also, how do you make it so the quations come up looking normal instead of typing it here, it's formatted different? Is that a website you type it in and link here? Thanks!

-Jeff
 
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Jeff231 said:

Homework Statement



A basketball player who is 2.00[m] tall is standing on the floor 10.0 [m] from the basket, as in the figure. If he shoots the ball at 40.0degrees angle with the horizontal, at what initial speed must he throw so that it goes through the hoob without striking the backboard? The basket height is 3.05[m].

Homework Equations



xf = xi + vit + 1/2at2
yf = yi + vit + 1/2at2

The Attempt at a Solution



I used the yf equation to come up with an equation that has vi and t as the variables. Then I used the xf equation and solved for t. Then I plugged in that t into the yf equation. It comes out to a pretty large quadratic equation:

-4.9 (10/vi(sin40))^2 + (vi(cos40))(10/vi(sin40)) - 1.05 = 0

Now I'm having trouble finding vi. I'm pretty sure I did everything right, but can someone help me figure out how to solve for vi?

Also, how do you make it so the quations come up looking normal instead of typing it here, it's formatted different? Is that a website you type it in and link here? Thanks!

-Jeff

First of all I hope you didn't show any acceleration in the x direction. I haven't checked what you did, but that term is 0 for Vx.

As to figuring combined x,y speed remember that the x and y components of speed add like vectors.

As to the superscripting and subscripting, I modified your Relevant Equations to show you how to code it. You can also use TEX (LaTex) but that is a more complicated tutorial. If you open up some of the posts with formulas, if you can make a 10 m jumpshot, that won't be so hard.
 
Last edited:

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