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Throwing a ball

  1. Sep 14, 2013 #1
    1. The problem statement, all variables and given/known data
    A boy throws a rock at speed v_0 at angle α from a balcony of height h. When the ball hits the ground, its velocity makes an angle θ with the ground. What is the minimum value of θ in terms of h and v_0


    2. Relevant equations

    basic kinematics equations.


    3. The attempt at a solution

    I started out with the following:

    [tex] v_x = v_0 \cos \alpha , v_y = \sqrt{v_0^2\sin^2 \alpha+2gh} [/tex]

    Then:

    [tex] f( \theta ) = \tan^{-1} (\frac{\sqrt{v_0^2 \sin^2 \alpha+2gh}}{v_0\cos\alpha} [/tex]

    To find the minimum, you have to differentiate, although whenever I do this i get a HUGE expression which doesn't make sense. Any help?Thanks :)
     

    Attached Files:

    Last edited: Sep 14, 2013
  2. jcsd
  3. Sep 14, 2013 #2
    When ## \theta ## is max, ## \tan \theta ## is max as well.
     
  4. Sep 14, 2013 #3
    But we need to minimize theta?
    Can you please give another hint?
     
  5. Sep 14, 2013 #4
    Ah, sorry. I really meant to say min, not max :)
     
  6. Sep 14, 2013 #5
    Gotcha.

    So am I correct in saying that after taking the tangent of both sides, the RHS equals zero? in which case :

    [tex] v_0^2 \sin^2 \alpha = 2gh [/tex] ??

    I attempted it again, however i cannot get rid of alpha here.
     
  7. Sep 14, 2013 #6
    You need ## (\tan \theta)' = 0 ##.
     
  8. Sep 14, 2013 #7
    I still don't get it. Why is that so? Thanks.
     
  9. Sep 14, 2013 #8
    Let ## f(x) ## be a monotonic function. Let ## g(x) ## be any function. Consider ## F(x) = f(g(x)) ##. ## F'(x) = (f(g(x)))' = f'(g(x)) g'(x) = 0 ##. Because ## f(x) ## is monotonic, ## f'(x) \ne 0 ##, thus the previous equation implies ## g'(x) = 0 ##, which means that if ## F(x) ## has a stationary point, ## g(x) ## has a stationary point there as well.

    ## \tan ## is monotonic from zero to ## \pi/2 ##.
     
  10. Sep 14, 2013 #9
    So does this mean I have to differentiate [tex] \frac{\sqrt{v_0^2\sin^2\alpha + 2gh}}{v_0\cos\alpha}[/tex]?
    Also, what do i differentiate with respect to? And how do i get rid of alpha? sorry I'm very stuck on this
     
  11. Sep 14, 2013 #10
    Since you are required to find that in terms of ##h## and ##v_0##, ##\alpha## has to disappear, which apparently means the minimum must be with respect to it.
     
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