Calculating Initial Speed for an Olympic Basketball Shot

AI Thread Summary
To calculate the initial speed for an Olympic basketball shot, the player must consider both horizontal and vertical components of motion. Key variables include the horizontal distance to the basket (5.64 m), the height of the basket (3.05 m), and the initial height of the ball (1.62 m), along with the launch angle (48 degrees) and gravitational acceleration (-9.81 m/s²). The discussion emphasizes the need to separate horizontal and vertical components to apply kinematic equations effectively. It is noted that the horizontal velocity remains constant, while vertical motion is influenced by gravity. Understanding these components is crucial to determine the initial speed required for the shot to score.
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Hi, I have a question that I can't finish, because it seems that I'm missing a particular variable. A little help would be very much appreciated.

Question: An Olympic basketball player shoots towards a basket that is 5.64 m horizontally from her and 3.05 m above the floor. The ball leaves her hand 1.62 m above the floor at an angle of 48.0 deg above the horizontal. What initial speed should she give the ball so that it reaches the basket and hopefully scores?

This is the variables that I found: xf: 5.64 m xi: 0m Yf:3.05 m yi: 1.62 m deg: 48 G: -9.81 m/s^2

Am I missing a variable, and if so can you tell me which one and explain how you founded it?
 
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I know that this is a two dimisional problem. But it seems like I don't have enough information to apply them to an equation of kineamatics. I need atleast three variables.

I only have two x values that I can use: they are: x: 5.64 m A: 0 m/s^2

I only have (three y vaules) technically two y values that I can use and they are: yi: 1.62m yf: 3.05 m G:-9.81
 
i don't think you're missing any variables. Basically you have the ball going up and coming down thru the basket and you have the ball traveling forward at speed vxi = vi * cos(48) and you know the upward speed is vyi = vi * sin(48).

so now use the equations you have and show some work.
 
Kineamatic Equations

V = Vo + at
X - Xo = Vot + .5at2
v2 = vo2 + 2a(X - Xo)
X - Xo = .5(Vo + V)t
 
What you have written doesn't mean anything because you haven't defined any of those things. The first thing you need to do is separate ""horizontal" and "vertical" components. What is the vertical acceleration. What is the horizontal acceleration?
 
I tried to get vxi from the equation x=xo+voxT+1/2aT^2

then I substitued the value into the same equation and got 5.64 m= 5.64 m.

I am really confused because I was expecting to get a time value. But I got the times canceling out and leaving me with meters=meters instead.
 
HallsofIvy said:
What you have written doesn't mean anything because you haven't defined any of those things. The first thing you need to do is separate ""horizontal" and "vertical" components. What is the vertical acceleration. What is the horizontal acceleration?

vertical acceleration: 9.81 m/s^2 (change to negative value because if gravity and velocity are both positive the ball woudn't come back down), so therefore the gravity will be -9.81 m/s^2.

horizontal velocity: is zero because air resistance is neglected and this means that it will travel at a constant velocity. Because there isn't no oxygen,carbon,... atoms causing the project tile to declerate.
 
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