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anonymous

  • one year ago

help

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  1. anonymous
    • one year ago
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    @xapproachesinfinity

  2. anonymous
    • one year ago
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    @Michele_Laino

  3. anonymous
    • one year ago
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    whats your question?

  4. anonymous
    • one year ago
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    The function H(t) = -16t2 + vt + s shows the height H (t), in feet, of a projectile launched vertically from s feet above the ground after t seconds. The initial speed of the projectile is v feet per second. Part A: The projectile was launched from a height of 100 feet with an initial velocity of 60 feet per second. Create an equation to find the time taken by the projectile to fall on the ground. (2 points) Part B: What is the maximum height that the projectile will reach? Show your work. (2 points) Part C: Another object moves in the air along the path of g(t) = 20 + 38.7t where g(t) is the height, in feet, of the object from the ground at time t seconds. Use a table to find the approximate solution to the equation H(t) = g(t), and explain what the solution represents in the context of the problem? [Use the function H(t) obtained in Part A, and estimate using integer values] (4 points) Part D: Do H(t) and g(t) intersect when the projectile is going up or down, and how do you know? (2 points)

  5. anonymous
    • one year ago
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    i wish i could help with this one but im not sure sorry :(

  6. anonymous
    • one year ago
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    thanks for trying

  7. Michele_Laino
    • one year ago
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    using your initial conditions, we can rewrite your equation, like below: \[\Large H\left( t \right) = - 16{t^2} + 60t + 100\]

  8. Michele_Laino
    • one year ago
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    furthermore the equation for speed v(t) is: \[\Large v\left( t \right) = 60 - 32t\]

  9. Michele_Laino
    • one year ago
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    when our projectile is going up, it will reach the maximum height, and at that time its speed is zero, so we can write this: \[\Large 0 = 60 - 32{t_1}\] where t_1 is the time needed to projectile in order to reach the maximum height

  10. Michele_Laino
    • one year ago
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    solving that equation for t_1 we have: \[\Large {t_1} = \frac{{60}}{{32}} = ...\sec \]

  11. Michele_Laino
    • one year ago
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    please complete that computation

  12. Michele_Laino
    • one year ago
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    please wait...

  13. anonymous
    • one year ago
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    no problem you are a lifesaver

  14. Michele_Laino
    • one year ago
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    ok! I'm here

  15. Michele_Laino
    • one year ago
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    we get: \[\Large {t_1} = \frac{{60}}{{32}} = 1.875\sec \] am I right?

  16. anonymous
    • one year ago
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    sure

  17. Michele_Laino
    • one year ago
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    now the maximum height reached by our projectile, is: \[\Large {H_1} = - 16 \cdot {1.875^2} + 60 \cdot 1.875 + 100 = ...meters\] please complete

  18. Michele_Laino
    • one year ago
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    what is H_1?

  19. anonymous
    • one year ago
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    idk

  20. Michele_Laino
    • one year ago
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    try to compute it, using a calculator, for example

  21. anonymous
    • one year ago
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    156.25

  22. Michele_Laino
    • one year ago
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    correct! \[\large H\left( t \right) = - 16 \cdot 1.8752 + 60 \cdot 1.875 + 100 = 156.25\;meters\]

  23. anonymous
    • one year ago
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    Awesome is that it am I done????

  24. Michele_Laino
    • one year ago
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    oops.. \[\large {H_1} = - 16 \cdot {1.875^2} + 60 \cdot 1.875 + 100 = 156.25\;meters\]

  25. Michele_Laino
    • one year ago
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    now, we have to compute the time needed to our projectile, in order to reach the ground, starting from its position at H_1 with speed=0

  26. Michele_Laino
    • one year ago
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    we have to use this equation: \[\Large H\left( t \right) = - 16{t^2} + 156.25\]

  27. Michele_Laino
    • one year ago
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    If I call with t_2 the time needed to our projectile in order to reach the earth surface, then we can write: \[\Large 0 = - 16t_2^2 + 156.25\] sinc qt t_2 H(t)=0 so the time t_2 is: \[\Large {t_2} = \sqrt {\frac{{156.25}}{{15}}} = ...\sec \] please compute that time t_2

  28. Michele_Laino
    • one year ago
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    oops.. since at t_2 H(t)=0...

  29. Michele_Laino
    • one year ago
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    oops.. another typo, here is the right formula: \[\Large {t_2} = \sqrt {\frac{{156.25}}{{16}}} = ...\sec \]

  30. Michele_Laino
    • one year ago
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    what is t_2?

  31. anonymous
    • one year ago
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    3.125

  32. Michele_Laino
    • one year ago
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    correct! So total time, needed to our projectile, is: \[\Large T = {t_1} + {t_2} = 1.875 + 3.125 = ...\sec \]

  33. anonymous
    • one year ago
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    5seconds

  34. Michele_Laino
    • one year ago
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    ok! more formally, if I call with h_0 the initial height, namely h_0=100 feet, and v_0 the initial speed, namely v_0=60 feet/sec, then time t_1 is: \[\large {t_1} = \frac{{{v_0}}}{g}\] as you can check from the above computations, furthermore, the maximum height reached by our projectile is: \[\large {H_1} = \frac{{v_0^2}}{{2g}} + {h_0}\] as you can check from the above computations again

  35. Michele_Laino
    • one year ago
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    so time t_2 is: \[\large {t_2} = \sqrt {\frac{{2{H_1}}}{g}} = \sqrt {\frac{2}{g}\left( {\frac{{v_0^2}}{{2g}} + {h_0}} \right)} \]

  36. Michele_Laino
    • one year ago
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    and total time is: \[\large T = {t_1} + {t_2} = \frac{{{v_0}}}{g} + \sqrt {\frac{2}{g}\left( {\frac{{v_0^2}}{{2g}} + {h_0}} \right)} \] which represents the requested formula

  37. Michele_Laino
    • one year ago
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    please, tell me when I may continue

  38. Michele_Laino
    • one year ago
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    of course, g is gravity, namely g=32 feet/sec^2

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