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Thread: 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
    $\displaystyle 2usinA-2g=0$

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

    Equation 2
    $\displaystyle 4ucosA=200$

    Sub 1 in 2

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

    $\displaystyle tanA=\frac{8g}{400}$

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

    is everything correct?

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

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

    The time to impact is given by

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

    Finally, you're going to have that

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

    Solve these equations for $\displaystyle v_{i}$.
    Last edited by Ackbeet; Jun 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

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

    The time to impact is given by

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

    Finally, you're going to have that

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

    Solve these equations for $\displaystyle v_{i}$.
    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:

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

    $\displaystyle v_{i}$ 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.

    $\displaystyle g$ is the acceleration due to gravity.

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

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

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

    There's a notational error in my equations:

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

    you know that

    $\displaystyle s = ut + \frac12 at^2$

    Let v be the initial velocity.

    Taking along the vertical;

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

    $\displaystyle 0 = 4vsin\theta - 78.4$

    $\displaystyle v sin\theta = 19.6$

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

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

    $\displaystyle 200 = 4vcos \theta$

    $\displaystyle v cos\theta = 50$

    Solve for v by eliminating the angle of projection theta.
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