Basketball Problem: Initial Speed to Make Shot

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A basketball player needs to determine the initial speed required to make a shot from a distance of 8.0 m at a 30.0° angle, with the basket height at 3.05 m and the player's height at 2.00 m. Initial calculations using projectile motion equations led to confusion over the correct values for sine and cosine, impacting the results. The player misapplied the equations for horizontal and vertical motion, particularly in how to relate time and distance. Suggestions were made to reassess the equations, especially the horizontal distance equation, to simplify the problem. A fresh approach is recommended to clarify the calculations and ensure accuracy in determining the initial speed.
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Homework Statement



A basketball player who is 2.00 m tall is standing on the floor L = 8.0 m from the basket, as in Figure P4.54. If he shoots the ball at a 30.0° angle with the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the backboard? The basket height is 3.05 m.


Homework Equations



y=Yo+Voyt-(g/2)t squared
X=Xo+Voxt-(g/2)t squared


The Attempt at a Solution



Vxo=cos(30)
Vyo=sin(30)
Vo (cos 30)=8

3.05=2.00+Vo-4.9t squared
3.05=10-4.9t squared
t=square root (6.95/4.9)
=1.191

Vo=8/(.7071*1.191)
=9.499


However, this is wrong...any suggesions??
 
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first check your cos 30 value.
 
oops...the the cosign value of 30 is .8660254038. So I take that value and multiply it by the square root of 6.95/4.9. Which is 1.191. I divide 8 by that product. Correct? My answer comes out to be 7.7561749; however, that is appearantly wrong. What am I doing wrong?
 
your sin value gets bollixed as well and the Vot (first order term) disappears--2Vo becomes 10. There may be more issues but start with those. Careful algebra for starts.
John
 
I have 2 Vo becoming 10 in my equation. I have:
3.05=10-4.9t squared
 
think we'are cross threaded, maybe I'm missing something
but,

y=Yo+sin(30)Vo*t+1/2at^2 going to 3.05=10-4.9t squared has me confused
 
bc 3.05 was the height of the basket. I set the equation :Yo+sin(30)Vo*t+1/2at^2 equal to the height of the basket. I am pretty confused. I'm not sure what I'm doing at this point:rolleyes: I have my final equation looking like Vo=8/(cosign30 * 1.191). I found the 1.191 by taking the square root of 6.95/4.9...AHHH!
 
Lets try a fresh start.

first assuming a conventional x/y horizontal and vertical coordinate system:

your eqn for Y is correct, x is not. Since there is no acceleration in in this direction, its simply
x=cos(30)Vo*t. But we do know how far the ball has to travel 8m.
so 8/cos(30)=Vo*t Now can we use that info in the Y eqn to make life simpler? For sure and I think you were close, so patience, my friend. Beware tho i think there may still be one wrinkle ahead.
 
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