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Projectile motion

  1. Oct 6, 2010 #1
    1. The problem statement, all variables and given/known data

    So the goal is to reach a distance H in the x-direction, starting from a height H in the y-direction, and you need to minimize the speed, and find the smalest and angle.

    2. Relevant equations

    i did:

    dx = V cos(theta) *t and
    y = y_0 +V sin(theta)*t-1/2*g*t^2

    Vx = Vcos(theta) Vy = Vsin(theta)

    3. The attempt at a solution

    i set them equal to each other and solve for time

    but I am not quite sure and the minimization part.
    do I have to take the derivative with respect of theta and set it equal to 0?
    Last edited: Oct 6, 2010
  2. jcsd
  3. Oct 6, 2010 #2
    There is a t missing in your formula for dx.

    If you equate the two you will get the angle and speed when you hit the target as you are considering the reverse process this way.

    If you want to calculate your process you have to change your equations for dx and dy a bit and then set them to your final coordinates.

    To get the smallest and largest angle you have to solve your resulting equation for v in terms of theta. Then you will be able to read of values of theta beyond which no solution exists.
    To minimize the speed you should differentiate w.r.t. theta and set it to 0.
  4. Oct 6, 2010 #3
    so i correct the equation put it in the t.

    By solving for V i got

    V = (1/2)*g*t/(cos(theta)-sin(theta))

    if I differentiate w.r.t. theta and set it = to 0 I got theta = -pi/4

    is that right, how do i get the speed?
  5. Oct 6, 2010 #4
    You have two equations. Use one to eliminate t and solve the second for v. There should be only one v in your first formulas. You have
    [tex]v_x=v\cos\theta[/tex] so you included the cos doubly.
    Last edited: Oct 6, 2010
  6. Oct 6, 2010 #5
    so just to understand, I solve for t and for the x equation and plug into the other one, so that I can eliminate t and I got.

    v = sqrt((1/2)*sqrt(2)*sqrt((2*sin(theta)^4-2+3*cos(theta)^2)*cos(theta)*h*g*(sin(theta)+cos(theta)))/(2*sin(theta)^4-2+3*cos(theta)^2))

    now I take the derivative w.r.t. theta and set it = to zero.

    I try it but how don't get a value for V?
  7. Oct 6, 2010 #6
    Your formula for y is still not correct. With increasing time your projectile would go higher and higer, a bit unrealistic.
    At t=0. You should have y=H. This is the initial condition.
  8. Oct 6, 2010 #7
    about now...?
  9. Oct 6, 2010 #8
    Yes. Now you can set for y and dx the coordinates of your target. This will give you two equations.
    Then you can use one to eliminate t and the second to solve for v.
    More from my side tomorrow if you still need help.
  10. Oct 6, 2010 #9
    i try but i dont really know how to get the value of V out.
  11. Oct 7, 2010 #10
    Where is your problem?
    What are your two equations? Can you solve them for t?
    Do you get an expression for v depending on theta?

    If you post your calculations I can check for errors and tell you what to try next.
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