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Projectile Motion

  1. Oct 12, 2009 #1
    A gun fires a projectile toward a target that is 4.54 km distant at the same elevation as the gun. The gunnery officer notes that the projectile hit the target 27.5 s after firing.

    a)Find the angle of elevation of the gun barrel.
    b)Find the initial speed of the projectile.

    So
    x=4.54 km
    t = 27.5 s

    I am really confused on how to solve this problem. It seems there isn't enough information to solve the problem or either I am missing something simple. Any help to get part A started would be great.
     
  2. jcsd
  3. Oct 12, 2009 #2

    lanedance

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    Homework Helper

    you know gravity and that teh bullet hits at the same elevation,

    assuming no wind resistance, then the initial vertical velocity (upward) will be the same magnitude as the final velocity downward. use these facts to find initail vertical velocity

    you also should be able to find the horizontal velocity, thus yielding the angle
     
  4. Oct 12, 2009 #3
    hi
    i have a problem from an in class lab where i have to find the initial velocity of the ball. given is the horizontal and vertical displacement of the projectile.
    Dy(height)=100cm
    Dx(range)=200m
    0(theta)= 15degrees.
    Find the initial velocity.
     
  5. Oct 13, 2009 #4

    lanedance

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    hey helpmee welcome to pf ;)

    you should start your own post and attempt some work first, and its better to post as other people can help as well, though the question sounds pretty similar to above... any ideas
     
    Last edited: Oct 13, 2009
  6. Oct 13, 2009 #5
    Thank you lanedance. I am still a little confused on finding the angle of elevation. Like you said we know Vo = V. and gravity and the bullet hits at the same time but how would I use this information to determine the angle. The question wants us to determine the angle before finding the initial speed. Could someone possibly give me a push start on the third unknown I am looking for to find the angle?
     
  7. Oct 13, 2009 #6
    It helps to break the problem up into the x and y components. You know there is no acceleration in the x component, and there is gravity accelerating in the y component. Find the Vi and Vf in both components and then solve for the distance traveled in each component. If you have the Vi in the x component and Vi in the y component, you can find the angle the gun is shot at (tan-1(Viy/Vix)

    Hope that makes sense.
     
  8. Oct 13, 2009 #7
    ok let me just attempt this and maybe you can help me from there..

    so I need two equations one for X and one for Y

    so...

    x-x0 = v0t +(1/2)at^2 since acceleration in the x direction is 0 we have

    x = v0t so we can solve for v0 in the x direction???

    v0 = .165 m/s?

    So now what about the y component. I am not sure of an equation dealing with that.
     
  9. Oct 13, 2009 #8
    Almost, x = v0t is not entirely correct as you need the x component of v0 in the x direction. So multiply by cos of the angle. x = v0cos(angle)t, solve for v0cos(angle).

    In the y direction, y=v0sin(angle)t - (1/2)at^2 because the y direction will have gravity as an acceleration. Because you know y = 0 (the cannon hits the target on the ground), solve for v0sin(angle).

    Once you have both v0cos(angle) and v0sin(angle) you can divide the two and use the previous equation I gave with the tan-1(v0sin(angle)/v0cos(angle)) to find the angle of the trajectory
     
  10. Oct 13, 2009 #9
    Ok I think this makes sense now...

    so

    x = v0 cos(theta)t
    v0cos(theta) = .165

    y= v0sin(theta)t - (1/2)(9.8)(27.5)^2
    v0sin(theta) = 3705.625

    so therefore

    tan^-1 (3705.625/.165) = 89.9 degrees.

    Does this look correct?
     
  11. Oct 13, 2009 #10
    A gun fires a projectile toward a target that is 4.54 km distant at the same elevation as the gun. The gunnery officer notes that the projectile hit the target 27.5 s after firing.

    a)Find the angle of elevation of the gun barrel.
    b)Find the initial speed of the projectile.

    (Note: your x and y vectors and independent)

    So....

    (delta y) = Vot + 1/2at^2

    so...

    0 = Vo + 1/2(-9.8)(27.5s)

    so Vy = 134.75 m/s

    Vx = Dx / t
    Vx = 4540 m / 27.5 s
    Vx = 165.1 m/s

    so Rx = sqrt(Vx^2 + Vy^2)
    Rx = 213.1 m/s

    and you angle is sin-1 (134.75/213.1)
    your angle is 39.2 degrees
     
  12. Oct 13, 2009 #11
    Does it really make sense that he fires the gun nearly straight up?
    [tex] x = v_o cos \theta t[/tex]
    [tex] y = v_o sin \theta t -\frac{1}{2}gt^2[/tex]
    You need to find theta so solve the first equation for [tex]v_o[/tex] in terms of [tex]cos\theta[/tex] and plug that into the second equation and solve for theta. Make sure you substitute the correct values.
     
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