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Projectile question

  1. Jan 18, 2006 #1


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    the question gives you the distance the projectile goes and how high it was at its highest points - they want to know what the initial velocity was. can someone get me started on how to approach this question? thanks.
  2. jcsd
  3. Jan 18, 2006 #2


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    Homework Helper

    The velocity of the projectile changes along it's path. Therefore you need to get the initial x and y velocity components of the projectile. The x-component do not change and can be calculated if you had the flight time of the projectile (which you don't). The flight time is double the time for the projectile to reach its maximum height.
  4. Jan 18, 2006 #3


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    how do you figure out the initial y component of the velocity at the highest point? i konw that its velocity at the highest point is 0 and it has -9.8 for acceleration due to gravity but what else can i use to solve in the y direction?
  5. Jan 18, 2006 #4


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    If you know the maximum height something reaches, then you can figure out how long it took to get up there and how long it takes to fall back down. These two times should be equal and are half of the total trip time (assuming the object takes off and lands at the same height).

    If the object starts from rest,
    The change in height = ½ at^2,
    Where a is the acceleration due to gravity (-9.81 m/s^2) and t is the time.
    If you know the max height of the object, calculate how long it takes to fall back down to earth, then double it to get the total trip time.
    As long as we are on the subject of falling back down to earth, we also know that the objects final downward speed when hitting the ground will be the same as its initial upward speed when taking off. So to calculate the initial and final Y velocity,
    V = a*t, where a is again the gravitational acceleration and t is the time the object accelerates which you just found earlier.

    Now that you know the total trip time of the object, and you also know the X distanced traveled, you can compute its initial X velocity,
    v = distance / time

    Now if you want to go even further and get the initial velocity and the angle above the horizon it was pointed at, you can use the Pythagorean theorem,
    v = sqrt(v_x^2 + v_y^2),
    the angle above the horizon will be,
    arctan (v_y / v_x)
  6. Jan 23, 2006 #5


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    You can determine the initial y velocity component, [itex]v_{yo}[/itex], of the projectile by using the given height, [itex]h[/itex], and noting that the velocity of the projectile is momentarily zero at the top of its trajectory:
  7. Jan 23, 2006 #6
    Theta = tan^-1(2H/R) where R is range and H is maximum height.
    You can use this formula if only theta is asked. I added this because this formula was not mentioned by others.
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