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?

- anonymous

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- anonymous

Guys please Help me

- anonymous

@radar could you please help me?

- anonymous

@robtobey

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## More answers

- anonymous

take the derivative and plug in 3 for t

- anonymous

is that how you find the distance? @sourwing

- anonymous

no. velocity

- anonymous

WHT DO YOU MEAN BY DERIVATE?

- anonymous

@sourwing ?

- anonymous

Guys please help me

- anonymous

@chrisdbest

- Kash_TheSmartGuy

Answer to a similar question - http://www.algebralab.org/Word/Word.aspx?file=Algebra_MaxMinProjectiles.xml

- anonymous

they're similar but they talk about the maximums @Kash_TheSmartGuy

- Kash_TheSmartGuy

Oh ok.

- anonymous

do you know someone that could help me? @Kash_TheSmartGuy

- Kash_TheSmartGuy

Probably @jim_thompson5910 or @campbell_st

- anonymous

@jim_thompson5910 could you help me?

- jim_thompson5910

this is for calculus class right?

- anonymous

yes

- jim_thompson5910

have you learned about derivatives? or differentiation?

- anonymous

No I haven't, but my professor explained that it was related to the slope formula

- jim_thompson5910

that's the average velocity, but they want the instantaneous velocity here

- jim_thompson5910

it seems odd how he's asking about a topic you haven't learned about yet

- anonymous

Wait I checked my notes isn't f(x) lim x-->0 f(a-h)-f(a) / h

- jim_thompson5910

yes it is

- jim_thompson5910

that's the limit definition of the derivative

- anonymous

okay, but what could I first?

- anonymous

Do I plug the 3 in the A right

- jim_thompson5910

yes

- jim_thompson5910

what are f(3) and f(3+h) equal to

- 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

F(3) gives me 71
and F(a+h) gives me h^2+6h+9

- jim_thompson5910

f(3) is correct but what you got for f(3+h) is not correct

- anonymous

isn't -144-96h-16h^2

- 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

Use
f(3) = 71
f(3+h) = -16h^2-91h+71
to find
\[\Large \frac{f(3+h)-f(3)}{h}\]

- anonymous

okay i see my mistake
-91

- 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

at that instant in time, the object is going -91 ft/sec (ie it's falling at 91 ft/sec)

- 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

you're welcome, I'm glad it helped out

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