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ERoseM

  • one year ago

A projectile is launched upward from the ground at 60 m/s. a. How long will it take to reach its highest point? b.) How high does it go? c.) How long does it take to hit the ground?

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  1. Yahoo!
    • one year ago
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    Time = u sin 60 /g H = u^2 sin^2 60 /2g T = 2u sin 60/g

  2. ERoseM
    • one year ago
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    Thank you but how did you get time and what does the u stand for?

  3. ERoseM
    • one year ago
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    I'm sorry I don't understand the u?

  4. Yahoo!
    • one year ago
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    u = initial Velocity

  5. ERoseM
    • one year ago
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    Thank you but I'm still confused.

  6. wio
    • one year ago
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    Okay, we start out with the fundamental idea that everything falls at constant acceleration. We'll call this constant acceleration (with respect to time) \(a\). We'll plug it in for the real value later.

  7. wio
    • one year ago
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    The velocity is the going to be the anti derivative: \[ v(t) = \int a dt = at+C \]We need to find the constant of integration. Suppose the initial velocity is \(v_0\) (we'll plug it in later). \[ v(0) = v_0 = a(t) +C = C \]So our constant of integration is just the initial velocity \(v_0\). We now have the following equation: \[ v(t) = at + v_0 \]Which is enough to answer question part a

  8. wio
    • one year ago
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    a. How long will it take to reach its highest point? <-- Well, the highest point is going to be the maximum position. Remember to find the maximum of a function, you need to find the critical number (when the derivative equals 0). Since velocity is the derivative of position with respect to time, we just need to set it to 0 to find critical numbers: \[ 0 = at + v_0 \implies at = -v_0 \implies t = -v_0/a \]So the amount of time it will take is just the initial velocity divided by acceleration. Don't worry about that negative sign. Since initial velocity is upward and gravity is downward, the negatives will cancel out. We will get a positive time value.

  9. wio
    • one year ago
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    b.) How high does it go? <-- For this we need our position function. Again we find the anti-derivative.\[ x(t) = \int v(t)dt = \int at+v_0dt = \frac{1}{2}at^2+v_0t+C \]We have another constant of integration. Again we'll suppose the initial distance of \(x_0\). \[ x(0) = x_0 = \frac{1}2{}a(0)^2 + v_0(0) + C = C \]So once again, our constant of integration is just the initial value. \[ x(t) = \frac{1}{2}at^2+v_0t+x_0 \] We know it will reach it's highest value at \(t = -v_0/a\), so let's plug that in: \[ \begin{split} x(-v_0/a) &= \frac{1}{2}a(-v_0/a)^2+v_0(-v_0/a)+x_0 \\ &= \frac{1}{2}\frac{v_0^2}{a}-\frac{v_0^2}{a}+x_0 \\ &= -\frac{1}{2}\frac{v_0^2}{a}+x_0 \\ &= -\frac{v_0^2}{2a}+x_0 \\ \end{split} \] So we have our max distance being: \[ x_{max} = -\frac{v_0^2}{2a}+x_0 \]

  10. ERoseM
    • one year ago
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    Okay, I understand that.

  11. ERoseM
    • one year ago
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    Now how do I apply real numbers?

  12. ERoseM
    • one year ago
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    Like where does the sixty actually belong?

  13. wio
    • one year ago
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    c.) How long does it take to hit the ground? <-- So we will call the position of the ground \(x_g\). We need to solve for \(t\) given \(x(t) = x_g\)\[ x_g = \frac{1}{2}at^2 + v_0t +x_0 \implies 0 = \frac{1}{2}at^2 + v_0t +x_0 -x_g \]This is just the root of a quadratic equation. \[ \begin{split} t &= \frac{-(v_0)\pm \sqrt{(v_0)^2-4(\frac{1}{2}a))(x_0-x_g)}}{2(\frac{1}{2}a)} \\ &= \frac{-v_0\pm \sqrt{v_0^2-2a(x_0-x_g)}}{a} \end{split} \] This is giving us two times. The earlier time is just when the projectile left the ground, we want the later time, when it hits the ground. Since the \(\sqrt{\ }\) is going to yield a positive value, we set the \(\pm \) to \(+\) to get the later time. \[ t=\frac{-v_0 + \sqrt{v_0^2-2a(x_0-x_g)}}{a} \]

  14. wio
    • one year ago
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    Now we can plug in actual numbers. What we need are \[ a \\ v_0 \\ x_0 \\ x_g \]

  15. wio
    • one year ago
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    Let's say that upward means 'positive' position, and downward means 'negative' position. \(a\) is gravitational acceleration downward. \(-9.8m/s^2\) \(v_0\) is our initial velocity: \(60m/s\) \(x_0\) is our initial position: \(0m\) \(x_g\) is the position of the ground: \(0m\)

  16. wio
    • one year ago
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    Taking a second look... I think we should try both signs of \(\pm\) rather than just the \(+\), my assumption that that \(+\) version would be higher was incorrect.

  17. wio
    • one year ago
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    Recap: \[ \begin{array}{rcl} t_{top} &=& -v_0/a \\ x_{top} &=& -\frac{v_0^2}{2a}+x_0 \\ t_{ground} &=& \max\left(\frac{-v_0+ \sqrt{v_0^2-2a(x_0-x_g)}}{a}, \frac{-v_0- \sqrt{v_0^2-2a(x_0-x_g)}}{a} \right) \\ a &=& -9.8m/s^2 \\ v_0 &=& 60m/s \\ x_0 &=& 0m \\ x_g &=& 0m \end{array} \]

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