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Homework Help: Projectile motion question totally stuck

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data

    A cannon is arrange to fire projectiles, with initial speed V0, directly up the face of a hill of elevation angle alpha. At what angle from the horizontal should the cannon be aimed to obtain the maximum possible range R up the face of the hill?

    2. Relevant equations

    v = v0 + at?

    x = x0 + v0t + at^2 ?

    3. The attempt at a solution

    I have no idea what to do, please help my self esteem is totally shot .. I can't believe that I actually have no clue
  2. jcsd
  3. Oct 20, 2009 #2


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    Just begin as you would any projectile problem.
    Separate the initial velocity into horizontal and vertical parts. Then write two headings:
    Horizontal and Vertical. Decide in each case whether you have constant speed or accelerated motion and write the appropriate formula(s).
    You'll need some model for the hill - it is like a straight line on a y vs x graph so you know its equation.
  4. Oct 20, 2009 #3
    Hi, thanks for the reply - I tried doing something like..

    Vx0 = V0costheta

    Vy0 = V0sintheta

    Rsinalpha = V0sintheta - 1/2gt^2

    Rcosalpha = V0costheta

    thanks a lot
    Last edited: Oct 20, 2009
  5. Oct 20, 2009 #4
    Realign your coordinate system so that the positive x axis is pointed up the slope. The angle of the slope comes into play when you calculate the accelerations in the x and y axes.
  6. Oct 20, 2009 #5

    can I just consider the hill as a straight line, calculate the angle from there and then just add on alpha after? I thought of this but I wasn't sure..

  7. Oct 20, 2009 #6
    Isn't the max for any equation 45 degrees
  8. Oct 20, 2009 #7


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    45 degrees is the optimum angle to maximize the range on a flat surface. Throwing up a hill at angle alpha, the optimum angle will vary with alpha.

    emyt, I was thinking that a hill of angle alpha will have a slope of tan(alpha) so its equation will be y = mx = tan(alpha)*x. [1]

    For the horizontal part, constant speed, you'll have something like
    x = v*cos(θ)*t [2]
    For the vertical part, instead of "Rsinalpha = V0sintheta - 1/2gt^2 " I would just write y = v*sin(θ)*t - 1/2gt^2 in this notation. [3]
    Sub [1] into [2] or [3] so you are looking at the situation when the projectile hits the hill. Study the remaining two equations and see if you can see what angle gives the maximum x or y that you are looking for.
  9. Oct 20, 2009 #8
    hi, thanks for the reply:

    I thought that I was looking for rcosalpha and rsinalpha as my final x and y.. since those would be the coordinates at the end of the range?

  10. Oct 20, 2009 #9


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    Oh, I see it now! I was confused at the lack of x's and y's. You are saying the same thing I did and you have already got it down to two equations. Missing a couple of t's, though. It's x = v*t and
    y = v*t + .5*a*t^2.

    If you solve the easier one for t and sub in the other, you get a nice quadratic in r so you should be able to get its maximum either using a derivative or recalling a quadratic's max from high school math.
  11. Oct 20, 2009 #10
    well, to find a max range
    cant we use R=u2sin2[tex]\theta[/tex]/g
  12. Oct 20, 2009 #11

    yeah I got it, thanks - I was repulsed by the hideous substituting... it wasn't so bad

    thanks again
  13. Oct 20, 2009 #12


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    That's a powerful little formula, look416!
    But it is only for shooting on a horizontal surface and doesn't apply to this problem.

    Congrats, emyt!
  14. Oct 20, 2009 #13
    hi, I tried putting -4.9(rcosalpha/V0x costheta)^2 + v0sintheta(rcosalpha/V0x costheta) - Rsinalpha = 0 into the quadratic formula and it got really messy :S should I be doing it like this? there are too many unknowns..

    thank you
  15. Oct 20, 2009 #14
    oh i got it
    thx for clearing my confusion
  16. Oct 20, 2009 #15


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    I didn't get such a messy quadratic as you. Mine had no c term so I could common factor it. However, I ended up with a formula for tan θ that blows up when alpha equals zero, so I must have an error.

    Perhaps if we work together on it we can find each other's errors.
    You are supposed to take the lead. How about we use A for alpha and T for theta so it isn't quite so messy? Do you have
    r*cos(A) = v*cos(T)*t and r*sin(A) = V*sin(T)*t - .5*g*t^2
    as the starting point? What did you get next?
  17. Oct 20, 2009 #16
    hi, I had the same thing, when I solved for t I got rcos(A)/vcos(T)

    then I plugged it into the sin equation and subtracted V*sin(T) from both sides to equate it to zero so I could find rcos(A)/vcos(T) and I thought that if I could assign a number to that, I could solve for theta.. but it didn't work

    thanks for working with me
  18. Oct 20, 2009 #17


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    Okay so putting t = rcos(A)/vcos(T) into r*sin(A) = V*sin(T)*t - .5*g*t^2
    gives you
    r*sin(A) = V*sin(T)*rcos(A)/vcos(T) - .5*g*r^2/v^2*cos^2(A)/cos^2(T)
    Why now subtract V*sin(T) from both sides? Suggest we divide both sides by r and take the two terms that have no r to the left side.

    This is fun - let's keep going!
  19. Oct 20, 2009 #18
    I equated it to zero so I could try solving for rcos(A)/vcos(T), but I guess I could divide by r as well, I just thought it would work with that method from what you said before
    thanks I'm on it
  20. Oct 20, 2009 #19
    hi, what did you get? :S I'm doing so much expanding and whatnot it doesn't seem right
    Last edited: Oct 20, 2009
  21. Oct 20, 2009 #20


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    sin(A) = V*sin(T)*cos(A)/vcos(T) - .5*g*r/v^2*cos^2(A)/cos^2(T)
    after dividing by r.
    I'll take the r term to the left and the sin(A) to the right:
    .5*g*r/v^2*cos^2(A)/cos^2(T) = V*sin(T)*cos(A)/vcos(T) - sin(A)
    How would you solve that for r?
  22. Oct 20, 2009 #21
    oh right, thanks.. then you divide and multiply to isolate r..

    r = (V*sin(T)*cos(A)/vcos(T) - sin(A)*[v^2*cos^2(A)/cos^2(T)])/.5*g

    but V theta and alpha are all unknown?
  23. Oct 20, 2009 #22


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    We must consider that alpha (A) is known - the specified hill angle.
    Our job is to find the theta that makes r a maximum.
    To avoid the mistake I made last time, we should check at this stage to see if it makes sense when alpha is zero.
  24. Oct 20, 2009 #23
    If you rotate your coordinate plane (I believe):
    [tex]x = vocos(\theta - \alpha ) t - \frac{1}{2}gsin\alpha t^2[/tex]

    [tex]y = vosin(\theta - \alpha) t -\frac{1}{2}gcos\alpha t^2[/tex]

    If you don't rotate your coordinate plane:
    [tex] x = vocos\theta t [/tex]

    [tex] y = vosin\theta t - \frac{1}{2}gt^2[/tex]

    [tex] tan \alpha = \frac{y}{x}[/tex]

    [tex] R = \sqrt{x^2+y^2}[/tex]

    [tex]\theta = [/tex] Launch angle
    [tex]\alpha = [/tex] Angle of incline
    Last edited: Oct 20, 2009
  25. Oct 20, 2009 #24


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    Thanks, Jebus. Regret I don't understand your very first step, the
    v*cos(theta)*t. It seems to me that after you rotate, your v would no longer be v but something like v*cos(alpha).

    emyt, I'm not agreeing with your last step. I get
    .5*g*r/v^2*cos^2(A)/cos^2(T) = V*sin(T)*cos(A)/vcos(T) - sin(A)
    .5*g*r = V^2*sin(T)*cos^3(A)/cos^3(T) - v^2*sin(A)*cos^2(A)/cos^2(T)
    and then
    r = 2/g*V^2*sin(T)*cos^3(A)/cos^3(T) - 2/g*v^2*sin(A)*cos^2(A)/cos^2(T)
  26. Oct 20, 2009 #25
    I probably should've stated the variables. I'll edit that in :>
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