1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Delphi51

    User Avatar
    Homework Helper

    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..

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

    Delphi51

    User Avatar
    Homework Helper

    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?

    thanks
     
  10. Oct 20, 2009 #9

    Delphi51

    User Avatar
    Homework Helper

    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

    Delphi51

    User Avatar
    Homework Helper

    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

    Delphi51

    User Avatar
    Homework Helper

    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

    Delphi51

    User Avatar
    Homework Helper

    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

    Delphi51

    User Avatar
    Homework Helper

    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Projectile motion question totally stuck
Loading...