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How would I solve for an objects time in flight which is going off a

  1. Nov 8, 2004 #1
    How would I solve for an objects time in flight which is going off a ramp at 28 degrees at 35 m/s and is 52 meters from the ground?

    I solved for the vertical and horizontal velocity's but im not sure how to solve for the entire time. I tried using d=Vit + 1/2at^2 but then i had to solve using the quadratic formula, which im not sure even worked. Whats the easiest way to do this?
  2. jcsd
  3. Nov 8, 2004 #2
    i would consider this problem in two dimensions, and assume the only motion it has is in the x and y direction. the question asks for its time in flight, so which direction would you need and what would the corresponding kinematic equation for velocity be?
  4. Nov 8, 2004 #3
    ok, so i solve for the max hieght within the parabola it does at the top, then add it to the hieght of the ramp. Can i just use Vf=Vi+at now?
  5. Nov 8, 2004 #4
    sorry i had a few drinks after class. you want to know the total time, so lets change that equation (which is correct) and change it to the equation for position.

    the time your interested in is the exact moment it hits the ground. you know that position (which is different depending on how you set up your axis), the initial position, and the initial velocity.
  6. Nov 8, 2004 #5
    hmm, i still dont seem to get the correct answer. I found the total vertical distance by finding the hieght of the object relitive to the top of the cliff, then added that to the height of the cliff. Then i used Vf = Vi + at, where Vf = 0? What am i doing wrong lol
  7. Nov 8, 2004 #6
    lol i dont even know why i solved for the height cause that equation doesnt even use it haha
  8. Nov 8, 2004 #7
    what equation are you using?

    r(new) = r(init) + v(init)t + (1/2)at^2

    is what i came up with, i could be always be wrong though.
  9. Nov 8, 2004 #8
    ive never seen that equation before lol. Ive been using

    Vf = Vi + at, Vf² = Vi² + 2ad and d = Vit + 1/2at²

    How would you solve it using those?
  10. Nov 8, 2004 #9
    that would be the equation for the position i derived. it allows me to shift my referance frame as i desire. it's the same as d = V(i)t + (1/2)at^2, but it allows me to set whatever origin i want -- i could choose 0 to be the ground, or i could choose it to be the top of the ramp, i would just have to adjust d appropriately to solve for the time.

    using your equation, 0m would be the heigth as t approaches 0. in that case, the final position of the object would be what? you know the initial velocity, but you have to remember to use only the y component of that velocity. the only acceleration is due to gravity, so you know that as well.

    the only thing you don't know is time, which i imagine you'll have to use the quadratic formula to solve for.
  11. Nov 8, 2004 #10
    oh ok, i was just wondering if there was any way around using the quadractic formula. Thx
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