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Math Help - projectile motion problem

  1. #1
    knowthebird
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    projectile motion problem

    The problem...
    A pitched ball is hit by a batter at a 45 degree angle and just clears the outfield fence, 98 m away. Assume that the fence is at the same height as the pitch and find the velocity of the ball when it left the bat. Neglect air resistance.
    What I have got...

    98 m = Velocityinitial * Cos(45) * ( (2 * Velocityinitial * Sin(45) ) / 9.8)

    I'm not sure if this is the correct way to solve for Vi, or how to solve it if it is.

    I guess I assume no acceleration changes during the flight.

    I'm also confused about how I can get the initial velocity with out knowing how long the total flight took. It seems like the there should be direct/indirect? relationship between flight time and initial velocity, but I guess not for some odd reason.

    Thanks ahead for any and all help.


    EDIT:
    Or should it be...
    sqrt((98*9.8)/sin(2*45)) just saw some equation like that online, so thought I would post another version.

    2nd EDIT:
    Ok, I used the equation above and got 30.99032107, and that works when I plug it into the equation I had above. So if it works in that its probably correct. I don't know why, but I have an answer.
    Last edited by knowthebird; April 12th 2007 at 05:41 AM.
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  2. #2
    Forum Admin topsquark's Avatar
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    A pitched ball is hit by a batter at a 45 degree angle and just clears the outfield fence, 98 m away. Assume that the fence is at the same height as the pitch and find the velocity of the ball when it left the bat. Neglect air resistance.
    These are always fun, aren't they?

    First of all, NEVER take an equation from another problem and just assume it holds. You should know WHY the equation works.

    The first thing to do is set an origin and coordinate system. I'll set the origin to be where the ball leaves the bat and let +x be in the direction of the horizontal component of the initial velocity and let +y be upward.

    Now set up a table of some kind, indicating what we know:
    t0 = 0 s (The time the ball left the bat)
    x0 = 0 m
    y0 = 0 m
    (These indicate the initial position of the ball)
    v0x = v0*cos(45)
    v0y = v0*sin(45)
    (These indicate the components of the initial velocity, v0)

    x = 98 m
    y =0 m
    (These indicate the final position of the ball, at the fence.)

    ax = 0 m/s^2
    ay = -9.8 m/s^2
    (These indicate the components of the acceleration of the ball)

    The unknowns are t (the time the ball gets to the fence), vx and vy (the components of the velocity when the ball gets to the fence), and of course v0 (the initial speed of the ball.)

    We need an equation with v0 in it. We have eight of them to choose from:
    x = x0 + v0x*t + (1/2)ax*t^2
    vx = v0x + ax*t
    vx^2 = v0x^2 + 2ax*(x - x0)
    x = x0 + (1/2)(v0x + vx)t

    y = y0 + v0y*t + (1/2)ay*t^2
    vy = v0y + ay*t
    vy^2 = v0y^2 + 2ay*(y - y0)
    y= y0 + (1/2)(v0y + vy)t

    The first four equations really reduce to only one, since ax = 0 m/s^2:
    x = x0 + v0x*t + (1/2)ax*t^2
    with the condition that vx = v0x.

    So
    98 = v0*t*cos(45)

    Now, we've got two unknowns in this equation, so we need another equation with only these two unknowns. The first y equation suits the bill:
    y = y0 + v0y*t + (1/2)ay*t^2

    or
    0 = v0*t*sin(45) - 4.9t^2

    0 = v0*sin(45) - 4.9t

    So our two equations to solve simultaneously are:
    98 = v0*t*cos(45)
    0 = v0*sin(45) - 4.9t

    Solve the bottom equation for t:
    t = [sin(45)/4.9]*v0
    and insert this into the top equation:

    98 = v0*{[sin(45)/4.9]*v0}*cos(45)

    98 = [sin(45)*cos(45)]*[1/4.9]*v0^2

    So
    v0 = sqrt{(98*4.9)/(sin(45)*cos(45)} = 30.99 m/s
    (We choose only the + value for the square root since v0 in these equations is a speed, and thus has a positive value.)

    -Dan
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  3. #3
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    Hello, knowthebird!

    A pitched ball is hit by a batter at a 45░ angle
    and just clears the outfield fence, 98 m away.
    Assume that the fence is at the same height as the pitch.
    Find the velocity of the ball when it left the bat.

    There are two equations for projectile motion (horizontal and vertical displacement):

    . . x .= .(vĚcosθ)t . . . y .= .h
    o + (vĚsinθ)t - 4.9t▓

    where v is the initial velocity, h
    o is the initial height and θ is the angle of elevation.


    Since the initial height of the ball and the height of the fence are equal: h
    o = 0.
    . . . . . . . . . . . . . . . . . . . . . . . _
    Since θ = 45░, sinθ = cosθ = 1/√2

    . . . . . . . . . . . . . . . . . . . . . . . . _ . . . . . . . . . . . _
    The equations become: .x .= .(v/√2)t . . . y .= .(v/√2)t - 4.9t▓


    When is the ball at the height of the fence? .(When is y = 0 ?)
    . . . . . _ . . . . . . . . . . . . . . . . . . . . . . . _
    . . (v/√2)t - 4.9t▓ .= .0 . . t .= .v/(4.9√2) .
    (and t = 0)


    At that time, x = 98 m.
    . . - . . - . - ._ . - . - . - ._
    . . x .= .(v/√2) Ě v/(4.9√2) .= .98 . . v▓/9.8 .= .98 . . v▓ .= .960.4

    . . . . . . . . . . . . . . . . . . . . . . . . . . . . _____
    Therefore, the initial velocity is: .v .= .√960.4 .= .30.99032107 . .31 m/sec

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