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Physics, projectile motion problem.

  1. Jul 21, 2006 #1
    Hi, here is my question.

    A gun shoots bullets that leave the muzzle at 247 m/s. If a bullet is to hit a target 161.4 m away at the level of the muzzle, the gun must be aimed at a point above the target. (Neglect air resistance.)
    How far above the target is this point if the angle the gun makes is less than 45degrees?
    The additional question is, same question above but if angle is more than 45degrees.

    For projectile motion problems, I know once we know the angle, then we take sin(angle)v to find the velocity for y-coordinate system, then solve for distance or time..whatever. But for this question, it's so ambiguous, angle is less than or greater than.. i have no idea what to do, help me out.
  2. jcsd
  3. Jul 21, 2006 #2


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    You need to consider the x component of motion and the y component. The x component is just its velocity of 247m/s multiplied by the cosine of whatever angle since air resistance is neglected and the y-component is the 247m/s multiplied by the sin of whatever angle. Now when you set these equations up and solve them I presume there will be two angles as solutions 1 above 45 degrees and 1 below.

    Now all you have to consider is the acceleration due to gravity until it hits the target.
    Last edited: Jul 21, 2006
  4. Jul 21, 2006 #3


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    The key in these problems is to solve for the time of flight first. As Kurdt says, if you neglect the slowing effect of air resistance, the horizontal velocity is constant. So use that to figure how long it will take for the projectile to horizontally reach the target. Then, using gravity's acceleration effect on the vertical component, calculate how high the projectile will go, and what the initial angle needs to be to give you that initial vertical velocity.

    The question is being a bit kind to you actually. They are giving you a clue that there are two possible answers to the initial angle. Think about it. The farthest you can shoot is if you start at a 45 degree angle, and if you start higher or lower than 45 degrees, you will land your ordinance shorter than the max.
  5. Jul 21, 2006 #4
    thanks for you guys advice. But I'm still not sure...

    when we want to find the time, the bullet travels,
    we have to know the angle first to be able to solve for t. we know the distance in horizontal but since we don't know the angle, what velocity should I use? I mean,
    x coordiante velocity is 237*cos(angle). If I don't know the angle, I can't find t. Which angle do i have to set? or how can I solve for angle?
    Last edited: Jul 21, 2006
  6. Jul 22, 2006 #5


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    You can solve for t in terms of the angle.

    Given that the initial angle is [itex]\theta[/itex], then the equations of motion are:
    [tex]v_x= 247 cos(\theta)[/tex]
    [tex]v_y= -gt+ 247 sin(\theta)[/itex]

    [tex]x= 247 cos(\theta)t[/tex]
    [tex]y= -\frac{g}{2}t^2+ 237 sin(\theta)t[/tex]

    You want to hit a target at the same height as the gun so you want
    [tex]y= -\frac{g}{2}t^2+ 237 sin(\theta)t= 0[/tex]
    which gives you a quadratic equation for t in terms of [itex]\theta[/itex].
    The distance to the target is 161.4 m so you must also have
    [tex]x= 247 cos(\theta)t= 161.4[/tex]
    Plugging t as a function of [itex]\theta[/itex] into that gives you a single equation for [itex]\theta[/itex].

    Since the first equation is quadratic, it will have two solutions. One gives you the "less than 45 degrees" solution, the other the "larger than 45 degrees solution".
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