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Homework Help: Review Q #5 (Projectile Motion)

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data
    A basketball leaves the hands of a shooter 2.1 meters above the floor and soars towards the basket, which is at a horizontal distance of 3 meter from the shooter. If the basket is 2.6 meters above the floor, and the player shoots the ball at an angle of 45 degree above the horizontal, calculate the initial velocity the shooter must give the basketball in order to score a basket.

    Yo = 2.1m
    Xo = 3m

    Y = 2.6m
    Ao = 45 degrees above horizontal (?)

    2. Relevant equations
    Vx = VoCosAo
    Vy = VoSinAo - gt


    3. The attempt at a solution
    I don't understand what the X is supposed to be?
     
  2. jcsd
  3. Sep 21, 2010 #2
    Ok, so I've drawn the diagram for this.
    It looks like this is solved with trig?
    I get a right triangle with the top as 2.6m. and the lower 45 degree angle as 2.1m.
    So the height of the triangle would be 0.5m and the base would be 3m.

    Then, what does the hypotenuse equal?
     
  4. Sep 21, 2010 #3
    This is projectile motion. No it cannot be solved with trig. You have to use physics too ;)
     
  5. Sep 21, 2010 #4
    Hehe.
    What am I missing :( ?
    I have no idea where to go from there (I've been looking at this for 10 minutes straight), yet again.
     
  6. Sep 21, 2010 #5
    First you know that the horizontal distance x=Vo*cos(45)*t.
    And you know the vertical distance: y=Vo*sin(45)*t-(g/2)t^2.
    Now from the first equation take t and plug it into the second equation. You will get an equation for y in terms of x.
    You want that y=0.5 and x=3 right? as the total height difference between the point of throw and the height of the basket is (2.6-2.1=0.5).
     
  7. Sep 21, 2010 #6
    3 = Vo * cos45o * t
    t = 3/ (Vo * cos45o)

    0.5 = Vosin45o-(9.8/2)t2
    0.5 = 3 tan45 - (9.8/2)(9/cos245oVo2

    Vo2= [(3tan45o-0.5)/(9.8/2)]*(9/cos245o)

    Vo2= (2.5/(9.8/2) * (0.5/9)
    Vo2= 0.2834467
    V = (0.2834467)^1/2 ?

    That's...odd?
    I don't think that's right...
     
  8. Sep 21, 2010 #7
    You messed up the calculation. Check that you expressed the velocity correctly.
     
  9. Sep 21, 2010 #8
    Alrighty, looks like I had computational error.

    0.5 = 3 -(9.8*9/(Vo^2*cos^2(45))/2
    11.025/ [Vo^2cos^2(45)]= 2.5
    11.025/2.5 = Vo^2 * cos^2(45)
    4.41 = [Vo^2]/2
    8.82 = Vo^2
    2.9698 m/s = Vo

    how about that?
     
  10. Sep 21, 2010 #9
    That still looks wrong ;)
     
  11. Sep 21, 2010 #10
    What could be it!? :(
     
  12. Sep 21, 2010 #11
    Check from here again :
    0.5 = 3 -(9.8*9/(Vo^2*cos^2(45))/2
    and calculate step by step
     
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