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Calculate total time taken for an object thrown at an angle from a height

  1. Mar 2, 2012 #1
    1. The problem statement, all variables and given/known data
    A cannon ball is fired with a velocity of 30m/s at an angle of 35° from the top of a cliff. The height of the cliff is 50m. Calculate total time of flight of the cannon ball before it hits the ground.

    [itex] v_o=30m/s [/itex]
    [itex] \alpha=35° [/itex]
    [itex] g=10m/s^2 [/itex]


    2. Relevant equations

    The equation of the parabola expressing the orbit of the cannon ball:
    [itex] y= \frac {-gx^2}{2v_o^2cos^2\alpha} + tan(\alpha) x [/itex]


    3. The attempt at a solution

    Obviously, a part of the parabola is missing as the cannon ball is struck from the cliff. I chose the position of the cannon as the origin and put a coordinate system there.
    I have 2 ways of solving this problem:

    1/ Calculate the time taken for the cannon ball to reach the maximum height:

    [itex] t=\frac {v_osin(\alpha)}{g} ≈ 1.7 (s) [/itex]
    The maximum height is: [tex] h=\frac{v_o^2 \sin^2(\alpha)}{2g}≈15(m) [/tex]

    Then, pretending that the cannon ball is struck vertically from the height of 15+50=65(m), the time taken for it to hit the ground is:
    [tex] \sqrt{\frac{2.65}{g}}≈3.6(s) [/tex]
    So that the total time is about: 1.7+3.6=5.3(s)

    2/ I want to place the cannon at a position that allows me to see the full parabola. I worked out the quadratic equation of the parabola, and I also had the new horizontal range of the cannon ball if the cannon is at a new position. However, here is where it get stuck. I do not know the time taken for the cannon ball to get from the new position to the original position, also the new angle at which it was thrown, and the new velocity.

    Can anyone please help me with this?

    I really appreciate your help!

    Thanks in advance!
    LovePhys
     
  2. jcsd
  3. Mar 2, 2012 #2

    tiny-tim

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    Hi LovePhys! :smile:

    Your first method is fine, but slightly quicker might be to use the standard constant acceleration equations in the x and y directions (s = ut and s = ut + 1/2 at2) for the whole trajectory :wink:
    I don't really understand what you're trying to do here. :confused:

    I don't think any such method is feasible.
     
  4. Mar 4, 2012 #3
    Hello tiny-tim,

    Thank you very much for your help and sorry for the late response!

    I am sorry that I did not explain my second method really clearly.

    The cannon is 50m above the ground, so a part of the trajectory is missing (in contrast with the fact that we can see the whole parabola if the cannon had been put on the ground). My idea is that I want to move the cannon on the ground so that I can calculate the total time of flight of the cannon ball. However, when I decide to do this, I realize that I am missing the information about angle of striking, and the cannon ball's initial velocity, as well as the time taken to reach the original position on the cliff, which is 50m above the ground.

    I am struggling with this and hope that you can give me a helping hand.

    Thank you!
    LovePhys
     
  5. Mar 4, 2012 #4

    tiny-tim

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    the only way you can find these things is to solve the original question first!
     
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