anonymous
  • anonymous
If a ball is thrown into the air with a velocity of 80 ft/s, its height in feet after t seconds is given by s(t) = 80t - 16t 2 . It will be at maximum height when its instantaneous velocity is zero. Find its average velocity from the time it is thrown (t = 0) to the time it reaches its maximum height.
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[80t-16t ^{2}\]
anonymous
  • anonymous
so i have \[\frac{(80(2+h)-16(2+h)^{2}-(80(2)-16(2)^{2})}{ h }\]
anonymous
  • anonymous
when i work the problem this way i don't get the right answer.

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freckles
  • freckles
i assume you are finding the derivative you just forgot to write as h approaches 0 but why are you find the derivative at x=2?
freckles
  • freckles
oops at t=2
freckles
  • freckles
to find the critical numbers of s(t) you need to first find s'(t) then solve s'(t)=0 for t
anonymous
  • anonymous
that just confused me.I suppose to Find its average velocity from the time it is thrown (t = 0) to the time it reaches its maximum height.
freckles
  • freckles
right to find the max you have to differentiate the function s(t) and solve the equation s'(t)=0 for t
anonymous
  • anonymous
so instead of 2 i use 0
freckles
  • freckles
where are you getting two from? and no instead of 2 you use t
freckles
  • freckles
\[s'(t)=\lim_{h \rightarrow 0} \frac{s(t+h)-s(t)}{h}\]
anonymous
  • anonymous
i was trying to use web asisgn to help with this problem, thats where i got the from
anonymous
  • anonymous
so that will be \[\frac{ 80(t+h)-16(t+h)^{2}-(80(t)-16(t)^{2} }{ h }\]
freckles
  • freckles
as h goes to 0
anonymous
  • anonymous
right
freckles
  • freckles
then once you evaluate that limit you set equal to 0 and solve for t
freckles
  • freckles
this is also in your problem... max height is achieved when instantaneous velocity is 0 aka when the derivative of s(t) is 0 that is why we are solving s'(t)=0 for t
freckles
  • freckles
if you are unsure how to go about evaluating the above limit you could multiply everything out and then cancel by looking at first the addition/subtraction part in the numerator and then cancel out common factors this is one way
freckles
  • freckles
common factors share by numerator and denominator*
freckles
  • freckles
@klewis1 let me know if you are having trouble evaluating the limit?
anonymous
  • anonymous
so now i'm stuck because i got \[\lim_{h \rightarrow 0}80-16h+32t\]
anonymous
  • anonymous
@freckles
freckles
  • freckles
I think you are off by a sign it looks pretty good just evaluate the limit by pluggin in 0 for h now it should be 80-32t by the way that is why I say you are probably off by a sign
anonymous
  • anonymous
yeah but the answer is 40
anonymous
  • anonymous
i just don"t how i will get to 40 from that
freckles
  • freckles
\[s'(t)=\lim_{h \rightarrow 0} \frac{80(t+h)-16(t+h)^2-(80t-16t^2)}{h} \\ s'(t)=\lim_{h \rightarrow 0} \frac{80t +80h-16(t^2+2th+h^2)-80t+16t^2}{h} \\ s'(t)=\lim_{h \rightarrow 0} \frac{80h-32th-16h^2}{h} \\ s'(t)=\lim_{h \rightarrow 0} [80-32t-16h]=80-32t\] now you have to solve s'(t)=0 for t
freckles
  • freckles
then you are asked to evaluate the average velocity between t=0 and when you have max height
freckles
  • freckles
80-32t=0 when t=?
freckles
  • freckles
let's call that t=a the last step is to find the average velocity between t=0 and t=a \[\frac{s(a)-s(0)}{a-0}=\frac{s(a)-s(0)}{a}=\frac{[80a-16a^2]-[80(0)-16(0)^2]}{a}\]
freckles
  • freckles
and you will get 40 at the end of all of this
anonymous
  • anonymous
lol i hope so
anonymous
  • anonymous
so i got 48 for a
freckles
  • freckles
a should be 5/2
anonymous
  • anonymous
i suppose divide got it
freckles
  • freckles
\[80-32t=0 \\ 80=32t \\ \frac{80}{32}=t \\ \frac{10}{4}=t \text{ divided top and bot by 8} \\ \frac{5}{2}=t \text{ divided top and bot by } 2\]
freckles
  • freckles
now just plug in 5/2 into the average velocity formula above
anonymous
  • anonymous
now i plug that back in fot t
freckles
  • freckles
well I called this t, a
freckles
  • freckles
so you replace the a in that average velocity formula I wrote above
anonymous
  • anonymous
yes i got it now

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