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Parametrics word problems

  1. May 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the height of the ball after 3 seconds if the ball is hit with an initial velocity of 90ft per second at an angle of 60° from an initial height of 5 ft.

    Find the horizontal distance a ball travels after 3 seconds if the ball is hit with an initial velocity of 90 ft per second at an initial height of 5 ft.

    If a ball is hit with an initial velocity of 110 ft per second at an angle of 45° from an initial height of 2 ft, how far will the ball travel before it hits the ground? How long will it be in the air?

    2. Relevant equations

    x=(v0cos θ)t and y=h + (v0 sin θ)t-16t^2

    3. The attempt at a solution
    I have tried plugging in the numbers to the equations but they do not work.
    x=(90cos60)t y=5+(90sin60)t-16t^2
    x=(110cos45)t y=2+(110sin45)t-16t^2
     
  2. jcsd
  3. May 24, 2013 #2

    Ray Vickson

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    As far as I can make out you do have the correct formulas, so what, exactly does not work out? Show us the complete numbers. (The only thing that comes to mind is that you *might* be evaluating sin and cos using software that expects angles in radians, while you are using degrees.)
     
  4. May 24, 2013 #3
    I don't know how exactly to find horizontal distance but these are the numbers i came up with after entering the equation in the calculator


    x=(90cos60)t y=5+(90sin60)t-16t^2
    after 3 seconds: 4.02ft

    x=(110cos45)t y=2+(110sin45)t-16t^2

    How long the ball will be in the air: 339.273 seconds
    How far it traveled before hitting the ground: y=2+(110sin45)339.273-16(339.273)^2= -1809940.95
     
  5. May 24, 2013 #4

    Ray Vickson

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    You should write x=(90cos60)t, y=5+(90sin60)t-16t^2; these are two *separate* formulas.

    The horizontal distance is just x, so substitute t = 3 into the expression and evaluate.

    When you say "after 3 seconds: 4.02ft", exactly what are you calculating? Horizontal distance? Vertical height, or what? Anyway, I cannot get anything like 4.02 ft; I get much larger numbers.

    Please show the details of your work. Do not just tell us what your calculator shows; tell us your values of 90*cos(60) and 90*sin(60), etc. Without knowing what you are doing we have no way to help you overcome your problems!
     
  6. May 24, 2013 #5


    For calculating the height after 3 seconds, I plugged in the parametric x=(90cos60)t, y=5+(90sin60)t-16t^2 and evaluated for x=3 on the graph, which gave me y= 4.02ft. In addition, when I try to evaluate the horizontal distance by plugging in t=3 I get -257.15

    x=(90cos60)t, y=5+(90sin60)t-16t^2
    90cos60 = -85.71, 90sin60 = -27.43
    x= -85.71(3), y= 5+(-27.43)3-16(3)^2
    x= -257.13, y= -221.29

    Should I be solving the equation in degrees instead of radians?
     
  7. May 25, 2013 #6

    SteamKing

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    Your calculations are wrong because you are evaluating sin and cos in radians instead of degrees.

    cos 60 degrees = 0.5
    sin 60 degrees = 0.866

    You should at least know the sin and cos of 0, 30, 45, and 60, and 90 degrees by heart.

    It is mindless errors like these which cause you to spend extra time on your homework and could cause you to fail an exam.
     
  8. May 25, 2013 #7
    1.
    x= (90cos60)t, y= 5+(90sin60)t-16t^2
    x= 90(.5)t, y= 5+(90(.866))t-16t^2

    For horizontal distance t=3 so,
    x= 90(.5)3 → x= 135ft

    For the height of the ball after 3 seconds,
    y= 5+(77.94)3-16(3)^2 → y= 94.82ft

    2.
    x= (110cos45)t, y= 2+(110sin45)t-16t^2

    To find how long the ball would be in the air I plugged the parametric equation
    x= (110(.707))t, y= 5+(110(.707))t-16t^2 into the calculator and evaluated the table for t which gave me 4.92 seconds.

    To find the distance of how far the ball traveled, I used the same equation above and looked at the parametric where y=0 which gave me 383.064 ft.
     
  9. May 25, 2013 #8

    Ray Vickson

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    If you had paid any attention at all to what I told you before, you would have realized this error right away; it was something I mentioned in my very first response to you.
     
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