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Math Help - Projectile problem

  1. #1
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    Projectile problem

    Problem requires the following to be answered: 1) Range, 2) total flight time and 3) max altitude. I need someone to review my range answer and my total flight time answer. In addition, I need some help setting up my max altitude.

    The equation is (vcosθ)ti +[(vsinθ)t -1/2 gt^2]j
    To get range, I set y=0 and solved for t.
    y= (vsinθ)t -1/2 gt^2
    0= (vsinθ)t -1/2 gt^2
    1/2 gt^2 = (vsinθ)t
    gt^2= 2(vsinθ)t
    gt= 2(vsinθ)
    t= 2(vsinθ)/g TOTAL FLIGHT TIME
    Inserted t into x= vcosθt
    vcosθ * 2(vsinθ)/g
    2 sinθcosθv^2/g
    Sin2θv^2/g RANGE

    Need help on setting up max altitude
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  2. #2
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    Quote Originally Posted by tigermadness View Post
    Problem requires the following to be answered: 1) Range, 2) total flight time and 3) max altitude. I need someone to review my range answer and my total flight time answer. In addition, I need some help setting up my max altitude.

    The equation is (vcosθ)ti +[(vsinθ)t -1/2 gt^2]j
    To get range, I set y=0 and solved for t.
    y= (vsinθ)t -1/2 gt^2
    0= (vsinθ)t -1/2 gt^2
    1/2 gt^2 = (vsinθ)t
    gt^2= 2(vsinθ)t
    gt= 2(vsinθ)
    t= 2(vsinθ)/g TOTAL FLIGHT TIME
    Inserted t into x= vcosθt
    vcosθ * 2(vsinθ)/g
    2 sinθcosθv^2/g
    Sin2θv^2/g RANGE

    Need help on setting up max altitude
    Yes, your formula is correct. If t is the time of flight, then x=v_{ix}t \implies t=\frac{x}{v_{ix}}. Consider the vertical motion alone; when the flight is over and the object strikes the target, vertical displacement =0=v_{iy}t + \frac{1}{2} (-g)t^2. Solving this equation gives t=2v_{iy}/g. But t=x/v_{ix}, so

    \frac{x}{v_{ix}} = \frac{2v_{iy}}{g} or: x=\frac{2v_{ix}v_{iy}}{g} = \frac{2(v_i cos \theta)(v_i sin \theta)}{g}

    You can now simplify the numerator using the double angle formula

     x= \frac{v^2_{i} sin 2 \theta}{g}
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  3. #3
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    So both of my TOTAL FLIGHT TIME and range answers are correct???
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  4. #4
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    Hello, tigermadness!

    Problem requires the following to be answered: 1) Range, 2) total flight time and 3) max altitude.
    I need someone to review my range answer and my total flight time answer.
    In addition, I need some help setting up my max altitude.

    The equation is: . \vec s \;=\; (v\cos\theta)\vec i + \left[(v\sin\theta)t -\tfrac{1}{2}gt^2\right]\vec j


    To get range, I set y=0 and solved for t.

    . . (v\sin\theta)t - \tfrac{1}{2}gt^2 \:=\:0 \quad\Rightarrow\quad \tfrac{1}{2}gt^2 \:=\: (v\sin\theta)t \quad\Rightarrow\quad gt^2\:=\:2(v\sin\theta)t . \Rightarrow\quad gt \:=\:2(v\sin\theta)

    . . t \:=\: \frac{2v\sin\theta}{g} . . . Total flight time . . Yes!

    Inserted t into v\cos\theta)t" alt="x\:=\v\cos\theta)t" />

    . . v\cos\theta)\left(\frac{2v\sin\theta}{g}\right) \;=\; \frac{v^2(2\sin\theta\cos\theta)}{g} \;= \;\frac{v^2\sin2\theta}{g}" alt="x \:=\v\cos\theta)\left(\frac{2v\sin\theta}{g}\right) \;=\; \frac{v^2(2\sin\theta\cos\theta)}{g} \;= \;\frac{v^2\sin2\theta}{g}" /> . . . Range . . Right!


    Need help on setting up max altitude.

    The height function is: . v\sin\theta)t - \tfrac{1}{2}gt^2" alt="y \:=\v\sin\theta)t - \tfrac{1}{2}gt^2" />

    Its graph is a down-opening parabola: \cap

    Its maximum is at its vertex: . t \;=\;\frac{-b}{2a} \;=\;\frac{-v\sin\theta}{2(-\frac{1}{2}g)} \;=\;\frac{v\sin\theta}{g}


    Insert into y\!:\;\;(v\sin\theta)\left(\frac{v\sin\theta}{g}\r  ight) - \frac{1}{2}g\left(\frac{v\sin\theta}{g}\right)^2


    Simplify: . \frac{v^2\sin^2\theta}{g} - \frac{1}{2}\!\cdot\!\frac{v^2\sin^2\theta}{g} \;\;=\;\; \frac{v^2\sin^2\theta}{2g} . . . maximum altitude

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