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  1. #1
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    Question projectile motion

    A shell, fired from a cannon, has a muzzle speed of 800ft/s. The barrel makes an angle of 45 degree with the horizontal and the barrel opening is assumed to be at ground level.

    How far does the shell travel horizontally?
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  2. #2
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    Quote Originally Posted by Jenny20 View Post
    A shell, fired from a cannon, has a muzzle speed of 800ft/s. The barrel makes an angle of 45 degree with the horizontal and the barrel opening is assumed to be at ground level.

    How far does the shell travel horizontally?
    I always draw a picture.

    So we've got an origin where the shell leaves the cannon (I'll assume it's on the ground at a height of 0 ft.) I am defining +x to the right and +y upward. The initial velocity of the shell is 800 ft/s, at an angle of 45 degrees above the horizontal (to the right in my diagram.)

    Note that the shell has no accleration component in the x direction. Thus we know that:
    x = x0 + v0x*t

    Now, x0 = 0 ft, because the shell started at the origin. v0x = v0*cos(45) = 565.685 ft/s, so
    x = 565.685*t

    We don't know t.
    So let's talk about the motion in the y direction.
    y = y0 + v0y*t + (1/2)ay*t^2

    Again, y0 = 0 ft, because the shell started at the origin. v0y = v0*sin(45) = 565.685 ft/s, and ay = -32 ft/s^2. So:
    y = 565.685*t - 16*t^2

    We want the point where the shell hits the ground, so we know that y = 0:
    0 = 565.685*t - 16*t^2

    Now solve for t:
    0 = t(565.685 - 16*t)

    Thus t = 0 s or 565.685 - 16*t = 0. The t = 0 s solution is where the shell started, we don't care about this. Thus:
    565.685 - 16*t = 0

    t = (565.685/16) = 35.355 s.

    Going back to the x equation:
    x = 565.685*t = 565.685*35.355 = 20000 ft.

    -Dan
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  3. #3
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    Hello, Jenny!

    A shell, fired from a cannon, has a muzzle speed of 800ft/s.
    The barrel makes an angle of 45 degree with the horizontal;
    the barrel opening is at ground level.

    How far does the shell travel horizontally?

    You may be expected to know the "projectile equations".

    . . x .= .v(cos θ)t

    . . y .= .h + v(sin θ) - 16tē

    where: .v = initial speed, h = initial height, θ = angle of elevation.


    We are given: .v = 800, h = 0, T = 45°
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ._
    The equations are: . x .= .(800cos45°)t .= . 400√2 t
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . _
    . . . . . . . . . . . . . . .y .= .(800sin45°)t - 16tē .= .400√2 t - 15tē


    The shell stops when it strikes the ground: .y = 0
    . . . . . ._ . . . . . . . . . . . . . . . . . . . . . _
    . . 400√2 t - 16tē .= .0 . . . . 16t(25√2 - t) .= .0
    . . . . . . . . . . . . . . _
    Hence: .t .= .0, 25√2 seconds

    . . . . . . . . . . _ . . . . . . . . . _ . . . ._
    When t = 25√2: .x .= .(400√2)(25√2) .= .2000 ft

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