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Homework Help: 3 projectiles, same velocity, different distance, same destination

  1. Aug 30, 2012 #1
    1. The problem statement, all variables and given/known data
    So I currently have 3 projectiles, that I need to get to the same destination. AND HIT, AT THE SAME TIME The projectiles are different distances from the target destination. How will I go about to find the angle all of the projectiles should be fired?

    3. The attempt at a solution
    I was thinking, if I could fire the furthest projectile at the optimal degree (45) and get that velocity and time, I will use that velocity and time for all the others. But is it possible to get to the same destination, from a different distance using the same velocity and amount of time, and if so, please tell me how!

    Thank you very much for reading this!
  2. jcsd
  3. Aug 30, 2012 #2
    Is it also required that the projectiles be launched at the same time?
  4. Aug 30, 2012 #3
    Yes at the exact same time..
  5. Aug 30, 2012 #4
    For a given distance D and magnitude of velocity V, what is the total flight time T?
  6. Aug 30, 2012 #5


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    Not possible then.

    The time of flight of a projectile is determined by the vertical component of its initial velocity. [Do you know why?]
    Change the angle and you change that component [assuming the initial actual speed of the projectile is the same].
    So if you change the angle, you change the time of landing, if the start times are the same.
    If you don't change the angle, then the range of the three is the same.
  7. Aug 30, 2012 #6
    Ah, okay.. I thought so.

    I suspected the question of being wrong, until today when one of my fellow students told me they had done it.

    Thank you for taking the time to answer! :-)

    I think i will just do it so that all 3 the projectiles reach the destination at a different time then.
  8. Aug 30, 2012 #7
    It looks like you need to find out the exact text of the problem.
  9. Aug 30, 2012 #8
    The problem was,

    We have 3 watch towers.

    We needed to create a c# program that will triangulate the position of the tank (using only the sound of the tank's fire, called sound ranging) and then let the 3 watch towers aim and fire at the tank, resulting in their mortars to hit the tank at the exact same time and firing at the exact same time.

    Now the triangulation was tough, but I did it. But this last part does not seem physically possible to me, unless we were allowed the change of velocity. (which we are not)

    Do you agree Voko?
  10. Aug 30, 2012 #9
    Range: R =u2 2sinθcosθ/g
    We can target the destination with 3 options, 45°, θ and (90-θ).
  11. Aug 30, 2012 #10


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    I think, not the velocity, but the initial speed of the projectiles is the same, but the firing angle is different. Am I right?

  12. Aug 30, 2012 #11
    I will try that, thanks :-)

    You are correct yes..
  13. Aug 30, 2012 #12
    For any given distance and initial velocity, there are only two trajectories, "low" and "high" (except the max range case, when they are the same trajectory). Each trajectory has a particular flight time, which I asked you about. The "high" flight time is, obviously, greater than the "low" flight time, so it might seem possible to hit the target from different distances using different kinds of trajectories. If you answer the question, you will see that "high" flight times are always greater than "low" flight times independently of the distance if the initial velocity is constant, so it is impossible to hit the target from different distances with equal flight times no matter what trajectory you choose.

    What you can do is hit the target at the same time, but you will have to fire with delays.
  14. Aug 30, 2012 #13
    Okay, a set of answers i calculated now are:
    -Tower 1 is 300m from the tank
    -Tower 2 is 200m from the tank
    -Tower 3 is 150m from the tank

    I will use the initial speed of 150m/s (as that is the speed of a mortar)

    Now the furthest one will need a angle of 3.758 to reach 300m and its time of flight is then 2 seconds.

    Now if we move on to the next.. the 200m one, the angle can either be 2.50128 making a time of flight 1.33461 seconds OR a angle of 87.4987 which leads to a time of flight of 30.5519seconds... That is what puzzles me!
  15. Aug 30, 2012 #14
    @Voko Yes, I agree with you. It just doesn't seem to work out..
  16. Aug 30, 2012 #15
    For any specified initial velocity, there is some max range. The max range flight time is T. For any other distance within the max range, you have two flight times, Tlow and Thigh. And Tlow < T < Thigh. At zero distance Tlow = 0 and Thigh is is the maximum possible flight time (for the velocity given, determined by shooting the projectile vertically). As the distance gets greater, Tlow grows bigger, and Thigh gets smaller, until the distance is max and Tlow = T = Thigh.
  17. Aug 30, 2012 #16


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    If you are using mortars, you should use the condition people use with them, a high angle.
    What I man is: if you send a projectile at an angle 10o to the horizontal, and another at 80o to the horizontal [same initial speed] they will land the same distance away.

    The 10o scenario is typical of field artillery, the 80o scenario is typical of a mortar.

    That does mean that the 300m example you used should be fired at about 87o rather than the 3o-ish answer you gave.
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