Basketball Vector/Kinematics Question

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In summary, the conversation discusses how to calculate the range of initial speeds allowed for a basketball player to make a shot into a basket at a certain height and angle, with a specific accuracy requirement. The motion of the basketball is described as a projectile, and the conversation touches on resolving the velocity into vertical and horizontal components and using equations for displacement to solve for the initial velocity. The conversation also references the use of trigonometry and the Wikipedia page on projectile motion for further guidance.
  • #1

Homework Statement

A basketball leaves a player's hands at a height of 2.10metres above the floor. The basket is 2.60metres above the floor. The player likes to shoot the ball at a 38 degree angle. If the shot is made from a horizontal distance of 11.00metres and must be accurate to +/-0.22metres (horizontally), what is the range of initial speeds allowed to make the basket?

I really have no idea how to do this... :(
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  • #2
You must show what you know and think about the problem. What kind of motion does the basketball perform?

  • #3
Parabolic? I tried to separate it into two (horizontal and vertical) components so that I could then work out the actual velocity. And then tried to work each of them out individually. I don't even know if that's the right way to go about doing it because it feels like I am missing one of the variables. But anyway, I tried to work out the vertical velocity first but was missing a variable for each of the available equations, so in the end I just tried subbing in zero for final velocity and went from there but the answers were all wrong.

I'd just like some guidance...
  • #4
That is a good start. The basketball moves like a projectile, along a parabola and its motion can be decomposed into horizontal and vertical ones. And you know two points of that parabola: Make a picture and show where they are. Show your calculations.
The final velocity will not be zero.

  • #5
I know that it is 2.1 metres from the ground when it starts and the end-point is 2.6 metres from the ground and 11 metres away from the start.

Therefore: X0 = 0m, X1 = 11m, Y0 = 2.1m, Y = 2.6m

The ball is being released at a 38 degree angle, my thought process was to get either one of the vector components somehow by playing with the formulas and then using trigonometry to work out the initial velocity at the release point.

But that is out of the question for the horizontal vector component, as I do not have the time that the ball is in the air, so that leaves the vertical vector component:

I tried setting final velocity as zero, so that the equation of v^2 = v0^2 + 2a(Y - Y0) would give me 0 = v0^2 + 2 * -9.8 * (2.6 - 2.1) which gave me the square root for 9.8 which was 3.13 m/s. But I know that to be wrong now because the answers I've gotten by using that value in my further calculations were incorrect.
  • #6
The final velocity is not zero.
You do not have time, but you have two equations for velocity and two equations for the displacements, both for the horizontal and vertical motions. You can make the time cancel.

Resolve the velocity into vertical and horizontal components. How do they depend on time?

Write out the equation for the vertical and horizontal displacements in terms of time.

You might find this page useful:

  • #7
Thanks for your help.
  • #8
ditde said:
Thanks for your help.

You are welcome.


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