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Projectile motion maximum range problem

  1. Oct 9, 2013 #1
    1. The problem statement, all variables and given/known data

    A projectile is fired at 30 km/h at the angle θ in x direction of an x-y plane. Ignoring the effects of air resistance, find the maximum
    range and the value of θ that gives this range

    Part A: write down the initial horizontal and vertical components
    of the velocity

    Part B: Write two equations to give the horizontal and vertical positions of the
    projectile as functions of time in terms of the initial speed, and the initial
    angle .

    Part C: Use these equations to show that the time the projectile will remain
    airborne is given by [itex]t = \frac{2v}{g}sin\theta[/itex] .

    Part D: Hence show that the range is given by [itex]\frac{2v^2}{g}sin(\theta)cos(\theta)[/itex]

    Part E: Show, by differentiating, that the maximum horizontal distance will occur if fired at 45° and is given by [itex]x_{max}=\frac{v^2}{g}[/itex]


    2. Relevant equations
    [tex]v=u+at \\
    s=ut+\frac{1}{2}gt^2 \\[/tex]


    3. The attempt at a solution

    Part A:
    Horizontal - [itex] vcos\theta[/itex]
    Vertical - [itex]vsin\theta[/itex]

    Part B:
    Horizontal - [itex]vcos(\theta)t[/itex]
    Vertical - [itex]vsin(\theta)t+\frac{1}{2}gt^2 [/itex]

    Part C is where I am stuck. I understand it but when I done it before I did it a different way (i think).

    When I did it before (last year) I used v=at so t=v/a so [itex]t=\frac{vsin\theta}{g}[/itex] but as that only gives it until the max height (halfway through) it needs to be times by two and hence [itex] t=\frac{2v}{g}sin\theta[/itex]

    But the question says to use what I did in Part B and im not sure how that is done.

    Part D: that is straightforward, just substituting what is found as t in Part C into the horizontal bit of Part B.

    Part E: This part has my head spinning at the moment.
     
    Last edited: Oct 10, 2013
  2. jcsd
  3. Oct 9, 2013 #2

    arildno

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    On C:
    At the time of solution, T, y=0, therefore, T satisfies which equation?
     
  4. Oct 9, 2013 #3

    arildno

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    E:
    You are to find maximal range. The range is given as a function of the angle v.
    Hint:
    It is easiest to first rewrite the expression for the range using the trig identity sin(v)cos(v)=1/2sin(2v)
     
  5. Oct 9, 2013 #4
    If you divided your vertical position by your horizontal position what would you have? (Hint: its a trig function)
     
  6. Oct 9, 2013 #5
    OK I think I see what you mean, like this?

    [tex]
    0=vsin(\theta)t-\frac{1}{2}gt^2 \\
    \frac{1}{2}gt^2=vsin(\theta)t \\
    \frac{\frac{1}{2}gt^2}{t}=vsin(\theta) \\
    \frac{1}{2}gt=vsin(\theta) \\
    gt=2vsin(\theta) \\
    t=\frac{2vsin(\theta)}{g} \\
    t=\frac{2v}{g}sin(\theta)
    [/tex]
     
  7. Oct 9, 2013 #6

    arildno

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    That's absolutely right!

    General cautionary note:
    Note that when you divided by "t", you lost one solution, namely t=0 (that is of course, the INITIAL y-position, so it isn't really relevant here, but you should not make a habit to divide with a quantity that MIGHT be 0)
     
  8. Oct 9, 2013 #7
    [itex]vtan(\theta)t-\frac{1}{2}gt^2[/itex] ?
     
  9. Oct 9, 2013 #8
    I think I may have done it. Withoutusing that identity though so not sure how succint it is.

    [tex]
    \frac{2v^2}{g}sin\theta cos\theta \\ for 0<= \theta <=90
    [/tex]
    then using the product rule
    [tex]
    \frac{2v^2}{g}(-sin\theta sin\theta+cos\theta cos\theta)=0 \\
    \frac{2v^2}{g}(-sin^2\theta+cos^2\theta)=0 \\
    [/tex]
    dividing by the fraction on the left then gets
    [tex]
    -sin^2\theta+cos^2\theta=0 \\
    cos^2\theta=sin^2\theta [/tex]
    then divide both sides by the cos^2
    [tex]
    \frac{cos^2\theta}{cos^2\theta}=\frac{sin^2\theta}{cos^2\theta} \\
    1=(\frac{sin\theta}{cos\theta})^2 \\
    1=tan^2\theta \\
    1=tan\theta \\
    tan^{-1}(1)=45
    [/tex]
     
    Last edited: Oct 9, 2013
  10. Oct 9, 2013 #9
    Never mind, this equation doesn't seem to work. What I was trying to do was use sin(θ)=y/x. I now realize that the θ in this sin(θ) changes with time, while the angle θ is constant in the position equations.
     
  11. Oct 9, 2013 #10
    Ok so it does work actually but you would have to use sin([itex]\frac{dθ}{dt}[/itex])=[itex]\frac{y}{x}[/itex]. Since the angle you start with will be the same as the angle at the end, [itex]\frac{dθ}{dt}[/itex]=0 therefore sin([itex]\frac{dθ}{dt}[/itex])=0. So you end up with

    0=[itex]\frac{vsin(θ)t-(1/2)gt^{2}}{vcos(θ)t}[/itex]=tan(θ)-[itex]\frac{gt}{2vcos(θ)}[/itex]

    then

    [itex]\frac{gt}{2vcos(θ)}[/itex]=tan(θ)

    t=[itex]\frac{2v}{g}[/itex]sin(θ)

    Its a little more complicated but it still works.
     
  12. Oct 9, 2013 #11

    arildno

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    That's perfectly okay!
    But, rewriting:
    [tex]Range=\frac{v^{2}}{g}\sin(2\theta)[/tex]
    Differentiate, and set equal to zero:
    [tex]0=\frac{2v^{2}}{g}\cos(2\theta)[/tex]
    That is,
    [tex]\cos(2\theta)=0[/tex]
    from which the same result follows! :smile:
     
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