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Math Help - Projectiles

  1. #1
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    Projectiles

    A particle of Q is projected from A at ground level. After 4 seconds the particle strikes the ground again at B where AB = 200m. Calculate the initial velocity of Q.

    using A=theta
    Equation 1
    2usinA-2g=0

    u=\frac{2g}{2sinA}

    Equation 2
    4ucosA=200

    Sub 1 in 2

    4*\frac{2g}{2sinA}cosA=200

    tanA=\frac{8g}{400}

    then  A=11.1
    and u=50.95ms^{-1}

    is everything correct?

    thanks for your help!
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  2. #2
    A Plied Mathematician
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    Not sure where you're getting your equations from. The range of the projectile is given by

    r=\frac{v_{i}^{2}}{g}\,\sin(2\theta).

    The time to impact is given by

    t=\frac{2v_{iy}}{g}.

    Finally, you're going to have that

    v_{iy}=v_{i}\sin(\theta).

    Solve these equations for v_{i}.
    Last edited by Ackbeet; June 23rd 2010 at 06:18 AM.
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    Not sure where you're getting your equations from. The range of the projectile is given by

    r=\frac{v_{i}^{2}}{g}\,\sin(2\theta).

    The time to impact is given by

    t=\frac{2v_{iy}}{g}.

    Finally, you're going to have that

    v_{iy}=v_{0}\sin(\theta).

    Solve these equations for v_{i}.
    what do i do with the equations?
    i don't really understand. the notations.

    thanks
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  4. #4
    A Plied Mathematician
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    Ok. Here's an explanation of all symbols in these equations:

    r is the horizontal distance along the ground between the launch point and the striking point. Otherwise called the range.

    v_{i} is the magnitude of the initial velocity. Also called the initial speed of the projectile. This is a scalar quantity, and is equal to the magnitude of the initial velocity vector.

    g is the acceleration due to gravity.

    \theta is the angle of elevation of the projectile when launched. It's the angle between the horizontal and the initial velocity vector.

    t is the amount of time it takes the projectile to strike the ground after having been launched.

    v_{iy} is the component of the initial velocity in the y-direction.

    There's a notational error in my equations:

    v_{iy}=v_{i}\sin(\theta), not v_{iy}=v_{0}\sin(\theta). Sorry about that. I've fixed that. Does this help explain things?
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  5. #5
    MHF Contributor Unknown008's Avatar
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    If you want to work form the basic principles;

    you know that

    s = ut + \frac12 at^2

    Let v be the initial velocity.

    Taking along the vertical;

    0 = vsin\theta (4) - \frac12 9.8 (4)^2

    0 = 4vsin\theta - 78.4

    v sin\theta = 19.6

    Taking along the horizontal, there is no acceleration due to gravity:

    200 = vcos\theta (4) + \frac12 (0)(4)

    200 = 4vcos \theta

    v cos\theta = 50

    Solve for v by eliminating the angle of projection theta.
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