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anonymous

  • one year ago

A ball is thrown vertically upward from the top of a 200 foot tower, with an initial velocity of 5 ft/sec. Its position function is s(t) = –16t2 + 5t + 200. What is its velocity in ft/sec when t = 3 seconds?

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  1. anonymous
    • one year ago
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    Guys please Help me

  2. anonymous
    • one year ago
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    @radar could you please help me?

  3. anonymous
    • one year ago
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    @robtobey

  4. anonymous
    • one year ago
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    take the derivative and plug in 3 for t

  5. anonymous
    • one year ago
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    is that how you find the distance? @sourwing

  6. anonymous
    • one year ago
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    no. velocity

  7. anonymous
    • one year ago
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    WHT DO YOU MEAN BY DERIVATE?

  8. anonymous
    • one year ago
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    @sourwing ?

  9. anonymous
    • one year ago
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    Guys please help me

  10. anonymous
    • one year ago
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    @chrisdbest

  11. Kash_TheSmartGuy
    • one year ago
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    Answer to a similar question - http://www.algebralab.org/Word/Word.aspx?file=Algebra_MaxMinProjectiles.xml

  12. anonymous
    • one year ago
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    they're similar but they talk about the maximums @Kash_TheSmartGuy

  13. Kash_TheSmartGuy
    • one year ago
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    Oh ok.

  14. anonymous
    • one year ago
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    do you know someone that could help me? @Kash_TheSmartGuy

  15. Kash_TheSmartGuy
    • one year ago
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    Probably @jim_thompson5910 or @campbell_st

  16. anonymous
    • one year ago
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    @jim_thompson5910 could you help me?

  17. jim_thompson5910
    • one year ago
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    this is for calculus class right?

  18. anonymous
    • one year ago
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    yes

  19. jim_thompson5910
    • one year ago
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    have you learned about derivatives? or differentiation?

  20. anonymous
    • one year ago
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    No I haven't, but my professor explained that it was related to the slope formula

  21. jim_thompson5910
    • one year ago
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    that's the average velocity, but they want the instantaneous velocity here

  22. jim_thompson5910
    • one year ago
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    it seems odd how he's asking about a topic you haven't learned about yet

  23. anonymous
    • one year ago
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    Wait I checked my notes isn't f(x) lim x-->0 f(a-h)-f(a) / h

  24. jim_thompson5910
    • one year ago
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    yes it is

  25. jim_thompson5910
    • one year ago
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    that's the limit definition of the derivative

  26. anonymous
    • one year ago
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    okay, but what could I first?

  27. anonymous
    • one year ago
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    Do I plug the 3 in the A right

  28. jim_thompson5910
    • one year ago
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    yes

  29. jim_thompson5910
    • one year ago
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    what are f(3) and f(3+h) equal to

  30. jim_thompson5910
    • one year ago
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    btw that should be f(a+h) and not f(a-h) but I guess it doesn't matter if h goes to 0

  31. anonymous
    • one year ago
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    F(3) gives me 71 and F(a+h) gives me h^2+6h+9

  32. jim_thompson5910
    • one year ago
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    f(3) is correct but what you got for f(3+h) is not correct

  33. anonymous
    • one year ago
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    isn't -144-96h-16h^2

  34. jim_thompson5910
    • one year ago
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    f(t) = -16t^2+5t+200 f(3+h) = -16(3+h)^2+5(3+h)+200 f(3+h) = -16(9+6h + h^2)+5(3+h)+200 f(3+h) = -144-96h-16h^2+15+5h+200 f(3+h) = -16h^2-91h+71

  35. jim_thompson5910
    • one year ago
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    Use f(3) = 71 f(3+h) = -16h^2-91h+71 to find \[\Large \frac{f(3+h)-f(3)}{h}\]

  36. anonymous
    • one year ago
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    okay i see my mistake -91

  37. jim_thompson5910
    • one year ago
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    yep \[\Large \frac{f(3+h)-f(3)}{h}=\frac{-16h^2-91h+71-71}{h}\] \[\Large \frac{f(3+h)-f(3)}{h}=\frac{-16h^2-91h}{h}\] \[\Large \frac{f(3+h)-f(3)}{h}=\frac{h(-16h-91)}{h}\] \[\Large \frac{f(3+h)-f(3)}{h}=\frac{\cancel{h}(-16h-91)}{\cancel{h}}\] Then plug in h = 0 to get -16h-91 to turn into -91

  38. jim_thompson5910
    • one year ago
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    at that instant in time, the object is going -91 ft/sec (ie it's falling at 91 ft/sec)

  39. anonymous
    • one year ago
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    okay, OMG thank you very muchI will write this down about the instantaneous velocity formula. I really appreciate all your help, god will pay you in a way that you will feel blessed. thank you, I had to submit this today at 12 and I am ready, thank you so much

  40. jim_thompson5910
    • one year ago
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    you're welcome, I'm glad it helped out

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