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skeller2

a batting machine launches a ball straight up in the air at a velocity of 100 ft/s. the machine is 65 feet tall what is the maximum height the ball can attain how fast will it be moving at this point?

  • one year ago
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  1. vf321
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    Well, let's start from the top. It's important to be able to derive the equations you need for physics without relying on memorization. Do you know calculus?

    • one year ago
  2. skeller2
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    Yes

    • one year ago
  3. vf321
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    Good. Assume that the ball, at any given moment, has constant acceleration \(a\). First, find a formula for \(v(t)\), the vertical velocity of the ball at any given time.

    • one year ago
  4. skeller2
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    I know that it needs to involve a (which would be negative due to gravity) and x over a certain period of time t. So \[v(t)=a(\int\limits_{0}^{t}xdt)\]?

    • one year ago
  5. vf321
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    Well don't worry about what the value of \(a\) yet (although you are right in that it is negative and the gravitational acceleration constant, which is \(-g = -9.8 \text{m}/\text{s}^2\). Solving for \(v(t)\) should be easy. If the derivative (acceleration) is constant, then \(v(t) = a t + C\) right?

    • one year ago
  6. skeller2
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    Ugh! Yes. It has been so long since I have done any word problems like this and it frustrates me that I can't remember how to start them off. I am really trying to help my cousin with her homework :/

    • one year ago
  7. vf321
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    Ok. well, we still have on more time-integral left. since this isn't for you, I'll just give you the solution: \(y = y_0 + v_0 t + a t^2/2\) I trust you can find approp. values for the constants, right?

    • one year ago
  8. skeller2
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    \[y_{0}=65 ft v _{0}=100ft/s \]

    • one year ago
  9. skeller2
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    Thank you so much for the help. I am clearly too tired after my own exams to be helping anyone else.

    • one year ago
  10. vf321
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    Yes @ your values. Max height can be found using single-derivative optimization (find where \(v(t) = 0\)).

    • one year ago
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