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Trajectory Problem

  1. Oct 31, 2005 #1
    I am not getting the same answer as the text for this problem:
    A cannon is fired from the top of the castle wall. The cannonball is fired at 50 m/s at an angle of 30 degrees. A cannonball that was accidentally dropped hits the moat below in 5s.
    (a) How far from the castle wall does the cannonball hit the ground?
    (b) What is the ball's maximum height above the ground?

    Known: x(i) = t(i) = 0
    v(i) = 50 m/s
    v(ix) = (50m/s)(cos30) = 43m/s = v(fx)
    v(iy) = (50m/s)(sin30) = 25m/s
    a(x) = 0
    a(y) = -9.8 m/s^2
    y(f) = 0
    For (a), I need to solve for x(f)

    Step 1. Given t=1.5s for a cannonball to drop to the ground, I can calculate the height of the castle wall or y(i) = 1/2 (-9.8m/s^2) (1.5s)^2 = 11m

    Step 2. Using the above value, solve for t:
    y(f)=0=y(i) + v(iy)(t) + 1/2a(y)(t)^2
    -11 = 25(t) - 4.9(t)^2
    -11 = t[25 - 4.9t]
    t = 0 or let 25 - 4.9t = 0
    t = -25/-4.9 = 5.1s

    Step 3, calculate x(f) = x(i) + v(ix)(t) + 1/2a(x)(t)^2
    x(f) = 0 + 43(5.1) + 0 = 219 m

    The answer in the text says it should be 239 m. Any insight on what I have done wrong?

    How do I calculate (b)?

    Thanks and sorry for the long post.
     
  2. jcsd
  3. Nov 1, 2005 #2

    andrevdh

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    Homework Helper

    I think the problem is where you are solving for t. You have -11 on the lefthand side of the equation not zero!
     
  4. Nov 1, 2005 #3
    Answe of the (B)

    When the ball reaches its maximum height above the ground, its velocity became zero, so: (assuming that g=10)
    The acceleration in the y direction is constant and equal to “-g” so we can write:
    v(fy)*v(fy) – v(iy)*v(iy)=-2gh and we know that v(fy)=0 so
    h = v(iy)*v(iy)/2g=(25*25)/(2*10) = 31.25
    H=h+h(i) which h(i) is the height of castle wall
     
  5. Nov 2, 2005 #4
    How should I solve for t? I revised my Step 2 above and came up with t=11s or t=7.3s --> both answers put my x(f) value way over what the answer is supposed to be.
     
  6. Nov 2, 2005 #5

    andrevdh

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    Homework Helper

    The general solution of a quadratic formula of the form:
    [tex]ax^2+bx+c=0[/tex]
    is given by:
    [tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
    It is derived by completing the square of the above equation. In this case we need to solve for the times when the equation [itex]-4.9t^2+25t+11[/itex] is zero, therefore
    [tex]\begin{equation}\begin{split}a=-4.9 \\
    b = 25 \\
    c = 11
    \end{split}\end{equation}[/tex]
    in this case. Use the second equation above to solve for the times - I got 5.5 s
     
    Last edited: Nov 2, 2005
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