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Please help an old man with a question.

  1. Dec 27, 2007 #1
    I need help with this question and an explanation of how you came up with the answer. Please help me if you can. I'm in a nursing home and it is a question from Grandson. Thanks in advance for any help you can give me.

    Question -

    "A basketball player shoots the ball and scores the winning basket. He released the ball at a height of 8 feet with an initial speed of 20 feet per second. If the ball's initial velocity vector made an angle of 45 degrees with the horizontal vector in the direction of the basket and if you ignore atmospheric effects, how far from the basket did the player shoot?"
  2. jcsd
  3. Dec 27, 2007 #2


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    Dearly Missed

    You need to know the vertical position of the basket in order to solve this problem.
  4. Dec 27, 2007 #3
    I'll solve in terms of variables, then you can see how it works

    It's all based around the formula
    [tex]d = v_0t + \frac{1}{2}at^2[/tex]

    In the vertical direction

    [tex]h_f - h_i = (v_0\sin\theta) t + \frac{1}{2}gt^2[/tex]

    Where h_f is the final height (the height of the hoop)
    h_i is the initial height (8 feet)
    v_0 is the initial speed
    g is 9.8 m/s^2
    and t is time

    In the horizontal direction

    [tex]d = (v_0\cos\theta) t[/tex]
    since the acceleration in the horizontal direction is 0, the whole 2nd term is gone

    Now solve for t in the first equation and plug it into the second to get the horizontal distance. You will get two solutions for t in the first equation because the ball reaches the height of the hoop twice (once on the way up, and once on the way down). Make sure you select the correct solution.
  5. Dec 27, 2007 #4
    Jiminey Christmas guys. I don't know how to figure that out. :confused: But I did need to explain how to get the answer so I appreciate that. A basketball rim is 10 feet high. Can anyone give me the answer? :approve: Thanks for your help!
  6. Dec 27, 2007 #5


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    I'm guessing you don't want to solve a quadratic equation. So let's do a simplified version. Suppose the ball is launched from 10ft instead of 8ft. Acceleration due to gravity is 32ft/sec/sec. The vertical and horizontal component of the initial velocity are equal since the angle is 45 and are equal to 20ft/sec*sin(45)=14ft/sec. Since the ball loses 32ft/sec of vertical velocity every second, it will take 14/32=.44sec for it to stop moving vertically at the top of the trajectory. It will take an equal time to fall down to the basket. So thats .88sec of flight. The horizontal velocity is also 14ft/sec, so the total horizontal distance is .88sec*14ft/sec=12ft. The distance is somewhat less with the extra two feet of vertical distance.
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