anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Guys please Help me
anonymous
  • anonymous
@radar could you please help me?
anonymous
  • anonymous
@robtobey

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More answers

anonymous
  • anonymous
take the derivative and plug in 3 for t
anonymous
  • anonymous
is that how you find the distance? @sourwing
anonymous
  • anonymous
no. velocity
anonymous
  • anonymous
WHT DO YOU MEAN BY DERIVATE?
anonymous
  • anonymous
@sourwing ?
anonymous
  • anonymous
Guys please help me
anonymous
  • anonymous
@chrisdbest
Kash_TheSmartGuy
  • Kash_TheSmartGuy
Answer to a similar question - http://www.algebralab.org/Word/Word.aspx?file=Algebra_MaxMinProjectiles.xml
anonymous
  • anonymous
they're similar but they talk about the maximums @Kash_TheSmartGuy
Kash_TheSmartGuy
  • Kash_TheSmartGuy
Oh ok.
anonymous
  • anonymous
do you know someone that could help me? @Kash_TheSmartGuy
Kash_TheSmartGuy
  • Kash_TheSmartGuy
Probably @jim_thompson5910 or @campbell_st
anonymous
  • anonymous
@jim_thompson5910 could you help me?
jim_thompson5910
  • jim_thompson5910
this is for calculus class right?
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
have you learned about derivatives? or differentiation?
anonymous
  • anonymous
No I haven't, but my professor explained that it was related to the slope formula
jim_thompson5910
  • jim_thompson5910
that's the average velocity, but they want the instantaneous velocity here
jim_thompson5910
  • jim_thompson5910
it seems odd how he's asking about a topic you haven't learned about yet
anonymous
  • anonymous
Wait I checked my notes isn't f(x) lim x-->0 f(a-h)-f(a) / h
jim_thompson5910
  • jim_thompson5910
yes it is
jim_thompson5910
  • jim_thompson5910
that's the limit definition of the derivative
anonymous
  • anonymous
okay, but what could I first?
anonymous
  • anonymous
Do I plug the 3 in the A right
jim_thompson5910
  • jim_thompson5910
yes
jim_thompson5910
  • jim_thompson5910
what are f(3) and f(3+h) equal to
jim_thompson5910
  • jim_thompson5910
btw that should be f(a+h) and not f(a-h) but I guess it doesn't matter if h goes to 0
anonymous
  • anonymous
F(3) gives me 71 and F(a+h) gives me h^2+6h+9
jim_thompson5910
  • jim_thompson5910
f(3) is correct but what you got for f(3+h) is not correct
anonymous
  • anonymous
isn't -144-96h-16h^2
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
Use f(3) = 71 f(3+h) = -16h^2-91h+71 to find \[\Large \frac{f(3+h)-f(3)}{h}\]
anonymous
  • anonymous
okay i see my mistake -91
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
at that instant in time, the object is going -91 ft/sec (ie it's falling at 91 ft/sec)
anonymous
  • anonymous
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
jim_thompson5910
  • jim_thompson5910
you're welcome, I'm glad it helped out

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