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Homework Help: Finding Maximum Height of a Projectile

  1. Sep 3, 2012 #1
    1. The problem statement, all variables and given/known data

    A basketball player is awarded free-throws. The center of the basket is [itex]4.21m[/itex] from the foul line and [itex]3.05m[/itex] from the ground. On the first free throw he throws the ball at an angle of 35 degrees above the horizontal with a speed of [itex]v_0 = 4.88 m/s[/itex]. The ball is released [itex]1.83m[/itex] from the floor. What is the maximum height reached by the ball?

    2. Relevant equations

    [itex]x = x_0 + v_0 t + (1/2) a t^2[/itex]
    [itex]v^2 = v_0^2 + 2 a \Delta x[/itex]
    [itex]v = v_0 + a t[/itex]
    [itex]v_y^2 = v_y_o^2 + 2gh[/itex]

    3. The attempt at a solution

    Since we were given both [itex]v_x[/itex] and [itex]v_y[/itex], we wanted to find the distance straight from the release point of the ball to the center of the rim.

    [itex]3.05m - 1.83m = 1.22m[/itex], which is height from the player's release point to the center of the rim.

    Then, [itex]cosθ = {1.22}/{x}[/itex]

    Therefore, the distance is [itex]5.20m[/itex]

    (This information was only relevant for a later part of the question)


    [itex]v_y^2 = v_yo^2 + 2gh[/itex] where 'h' is height, which is what we needed to solve for.

    Solving for [itex]h[/itex], we find that

    [itex]h={-v_yo}/{2g}[/itex]. Then,

    [itex]h={-(4.88 sin35)^2}/{2(-9.8m/s^2)} ≈ 0.4m[/itex]

    Adding [itex]0.4m[/itex] to the basketball's original height of [itex]1.83m[/itex], and you get the ball's maximum height to be [itex]2.23m[/itex] above the ground.

    There is no soution for me to verify this answer with, so I would appreciate anyone telling me where I messed up, or in the less likely scenario, that I got the answer correct.
    Last edited: Sep 3, 2012
  2. jcsd
  3. Sep 4, 2012 #2


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    I don't see any error. This maximum height falls short of the height of the basket.
  4. Sep 5, 2012 #3
    Thanks for the response! A friend and I were working out some challenge problems (not much of a challenge, to be honest), but we have no answer key.

    The ball falling short of the basket actually works out for the next problem, which wants us to figure out at what velocity he needs to shoot the basketball in order to make it into the basket.

    As a side note, I apologize for any botched usage of LaTeX. Hopefully it didn't make this too painful to read.
  5. Sep 5, 2012 #4
    About your LaTeX. If you want to write [itex]v_{y_0}[/itex], then you should write v_{y_0}. So be sure to enclode y_0 in brackets. Just writing v_y_0 will be an error!
  6. Sep 6, 2012 #5


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    Are you needing help with this question? Using the standard kinematic relations; [tex] s_y = v_{oy}t - \frac{1}{2}gt^2 [/tex] and [itex] t = \frac{s_x}{v_{ox}}, [/itex] you can derive the well known trajectory equation; [tex] s_y = s_x\tan\theta - \frac{gs_x^2}{2v_o^2\,\cos^2\theta}. [/tex]
    Set up a suitable coordinate system and plug in to this equation the coordinates of the basket net and simply solve for [itex] v_o. [/itex]
  7. Sep 6, 2012 #6
    I'm a little confused with this equation.

    Take a problem that I have to solve (which is very similar to this one) which asks for the initial velocity in which the basketball needs to be shot in order for the player to make the shot.

    Now, when I isolate [itex]v_o^2[/itex], I get: [tex]v_o^2=\frac{S_xtanθ-gs_x^2}{S_y2cos^2θ}[/tex]

    Assuming that this equation is correct, I would need to know both the speeds of the x-axis, and the speed of the y-axis ([itex]S_x^2[/itex] and [itex]S_y^2[/itex]), in which case I wouldn't even need to use your derived equation, considering that I would then know both of the components of the resultant vector of [itex]v_o^2[/itex], which I could use to find that vector.

    I'm mainly having difficulty figuring out what this equation is useful for. (Not that I don't appreciate you sharing it with me).
  8. Sep 7, 2012 #7


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    Maybe my notation is a bit confusing. [itex] s_x [/itex] and [itex] s_y [/itex] here are like x and y. So the point [itex] (s_x , s_y) [/itex] represents some point on the trajectory. Setting up a coordinate system with the origin at the point of launch, you know the height of the basket net relative to your coordinate system [itex] (s_y) [/itex]and it's horizontal distance [itex] (s_x)[/itex].Then your only variable is [itex] v_o [/itex] in the trajectory equation.
    Just a quick remark: this is assuming you want to know [itex] v_o [/itex] at the given angle.
  9. Sep 7, 2012 #8
    Oh alright, that makes a lot more sense. Thanks!
  10. Sep 7, 2012 #9
    I don't mean to interject -- but I was wondering about the original post, if that is introductory physics what I am doing is like pre k physics? I don't want to steal any shine or more attention from the original question. Please contact me or quote me or something if you want to. What am I doing it seems like I am wasting my time trying to understand the material but if I don't start from page 1 where do you start?
  11. Sep 8, 2012 #10
    I'm in AP Physics, so the material might be a little more advanced than what you would see in a regular physics class.

    Not only that, but the question that I posted here is a "challenge" question that our teacher offered. None of the questions on the test were as hard as the challenge problems, so what you are seeing here is honestly as advanced as our teacher could make the questions (while still giving us a realistic chance of solving them).
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