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anonymous

  • 5 years ago

GIVEN: A projectile is fired from the bottom of a 240 ft. deep gorge and is visible only when the projectile is above the rim of the gorge. If the projectile is fired with an initial velocity of 175 ft/s, the height of the projectile from ground level after t seconds is given by s(t) = -16t2 + 175t - 240 (a) During what interval can the projectile be seen? (b) What is the maximum height of the projectile?

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  1. anonymous
    • 5 years ago
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    for b...you can find the maximum height by taking the derivative, solving for the zeros of the derivative and doing the first derivative test to find the maximum

  2. anonymous
    • 5 years ago
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    a) since the projectile can only be seen from above the rim of the gorge, \[-16t^2+175t-240\ge 240\] solve this and take only the positive values of t

  3. anonymous
    • 5 years ago
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    ^^ if any

  4. anonymous
    • 5 years ago
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    u should find that the maximum is when t=175/32

  5. anonymous
    • 5 years ago
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    I calculated a max height of 238.5, but does that mean it would always fall short of the 240 foot height rim of the gorge?

  6. anonymous
    • 5 years ago
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    yes

  7. anonymous
    • 5 years ago
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    So there is no interval during which the projectile would be visible at ground level, right?

  8. anonymous
    • 5 years ago
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    when you solve the equation given for part a greater than 240...it will never exist

  9. anonymous
    • 5 years ago
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    b) the maximum height will be at -32t+175=0 that is t=175/32 therefor the maximum height is \[H_\max= v_i+s(175/32)\]

  10. anonymous
    • 5 years ago
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    sorry

  11. anonymous
    • 5 years ago
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    wait...i just realized that the equation is given from ground level

  12. anonymous
    • 5 years ago
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    so the maximum height the is 238.51 above ground level

  13. anonymous
    • 5 years ago
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    so part a is written wrong

  14. anonymous
    • 5 years ago
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    −16t2+175t−240≥0

  15. anonymous
    • 5 years ago
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    you should be solving −16t2+175t−240≥0 for part a

  16. anonymous
    • 5 years ago
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    \[H_\max=V_i/t_o+s(t_0), t_0=175/32\]

  17. anonymous
    • 5 years ago
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    \[V_i*t_0\]*

  18. anonymous
    • 5 years ago
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    I am making so many mistakes :(

  19. anonymous
    • 5 years ago
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    for part a you should get t needs to be between 1.6078 and 9.3297

  20. anonymous
    • 5 years ago
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    dont worry....i misread the problem at first too

  21. anonymous
    • 5 years ago
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    But why isn't the max height then 478.5 (240+238.5), if its 238.5 feet above ground level. How can s(0) be at ground level when its being shot from 240 feet underground? I am so confused.

  22. anonymous
    • 5 years ago
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    its not above ground until 1.6078

  23. anonymous
    • 5 years ago
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    im guessing that part b is the max height from ground and not from the gorge since the equation given is from the ground

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