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Math Help - Parametric equation problem?

  1. #1
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    Parametric equation problem?

    The human cannonball is shot out of a cannon with an initial velocity of 70 mph 10 feet above the ground at an angle of 35 degrees.

    What is the maximum range of the cannon?

    I did this parametric equation
    x=t v cos
    y= t v sin-1/2gt^2+h

    So I set y to zero and to find t and I got t= 3.14
    Then I plug it in back to t in and got x=180.04
    As I checked the answer so I am not sure what I may have done wrong?

    These are the formulas I used
    x=t 70 cos (35)
    y= t 70 sin (35)-16t^2+10
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  2. #2
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    Hello, homeylova223!

    The human cannonball is shot out of a cannon with an initial velocity of 70 mph
    10 feet above the ground at an angle of 35 degrees.

    What is the maximum range of the cannon?


    \text{These are the formulas I used:}

    . . x \:=\:(70\cos35)t . no

    . . y \:=\:10 + (70\sin35)t - 16t^2 . no

    \text{I set }y\text{ to zero and to find }t\text{ and I got: } t= 3.14 . no

    "They" pulled a dirty trick on us!

    The height is in feet . . . the gravitational constant is in ft/sec ^2.

    But the velocity is in miles per hour!


    We find that 70 mi/hr is equal to \frac{308}{3} ft/sec.


    The equations are: . \begin{Bmatrix}x &=& \left(\frac{308}{3}\cos35\right)t \qquad\qquad\quad \\ \\[-3mm]<br />
y &=& 10 + \left(\frac{308}{3}\sin35\right)t - 16t^2 \end{Bmatrix}

    Now we can proceed . . .

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  3. #3
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    Excellent, Soroban, I missed that completely!
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    Thanks Soroban.
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  5. #5
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    I have one more question
    I set y to zero ended up getting t=4.305
    then plugged that back into x equation to get x=362.086 ft
    While apparently the correct answer is 323.2 ft
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  6. #6
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    Quote Originally Posted by homeylova223 View Post
    I have one more question
    I set y to zero ended up getting t=4.305
    then plugged that back into x equation to get x=362.086 ft
    While apparently the correct answer is 323.2 ft
    your time is incorrect.

    0 = 10 + \dfrac{308}{3} \sin(35) \cdot t - 16t^2

    a = -16 , b = \dfrac{308}{3} \sin(35) , c = 10

    t \approx 3.84 sec
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  7. #7
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    I am slightly confused because I did this

    0=10+308/3 sin 35-16t^2
    0=10+58.887t-16t^2 factor t
    0=68.887-16t
    16t=68.887
    t=4.305

    What error could have been made?
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  8. #8
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    Quote Originally Posted by homeylova223 View Post
    I am slightly confused because I did this

    0=10+308/3 sin 35-16t^2
    0=10+58.887t-16t^2 factor t
    0=68.887-16t
    16t=68.887
    t=4.305

    What error could have been made?
    10 + 58.887t \ne 68.887 t

    ... the quadratic will not factor. use the quadratic formula.
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