Question

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.

Answer 1

\[80t-16t ^{2}\]

Answer 2

so i have \[\frac{(80(2+h)-16(2+h)^{2}-(80(2)-16(2)^{2})}{ h }\]

Answer 3

when i work the problem this way i don't get the right answer.

Answer 4

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?

Answer 5

oops at t=2

Answer 6

to find the critical numbers of s(t)
you need to first find s'(t)
then solve s'(t)=0 for t

Answer 7

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.

Answer 8

right to find the max you have to differentiate the function s(t) and solve the equation s'(t)=0 for t

Answer 9

so instead of 2 i use 0

Answer 10

where are you getting two from?
and no instead of 2 you use t

Answer 11

\[s'(t)=\lim_{h \rightarrow 0} \frac{s(t+h)-s(t)}{h}\]

Answer 12

i was trying to use web asisgn to help with this problem, thats where i got the from

Answer 13

so that will be \[\frac{ 80(t+h)-16(t+h)^{2}-(80(t)-16(t)^{2} }{ h }\]

Answer 14

as h goes to 0

Answer 15

right

Answer 16

then once you evaluate that limit
you set equal to 0
and solve for t

Answer 17

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

Answer 18

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

Answer 19

common factors share by numerator and denominator*

Answer 20

@klewis1 let me know if you are having trouble evaluating the limit?

Answer 21

so now i'm stuck because i got \[\lim_{h \rightarrow 0}80-16h+32t\]

Answer 22

@freckles

Answer 23

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

Answer 24

yeah but the answer is 40

Answer 25

i just don"t how i will get to 40 from that

Answer 26

\[s'(t)=\lim_{h \rightarrow 0} \frac{80(t+h)-16(t+h)^2-(80t-16t^2)}{h} \\ s'(t)=\lim_{h \rightarrow 0} \frac{80t +80h-16(t^2+2th+h^2)-80t+16t^2}{h} \\ s'(t)=\lim_{h \rightarrow 0} \frac{80h-32th-16h^2}{h} \\ s'(t)=\lim_{h \rightarrow 0} [80-32t-16h]=80-32t\]
now you have to solve s'(t)=0 for t

Answer 27

then you are asked to evaluate the average velocity between t=0 and when you have max height

Answer 28

80-32t=0 when t=?

Answer 29

let's call that t=a

the last step is to find the average velocity between t=0 and t=a
\[\frac{s(a)-s(0)}{a-0}=\frac{s(a)-s(0)}{a}=\frac{[80a-16a^2]-[80(0)-16(0)^2]}{a}\]

Answer 30

and you will get 40 at the end of all of this

Answer 31

lol i hope so

Answer 32

so i got 48 for a

Answer 33

a should be 5/2

Answer 34

i suppose divide got it

Answer 35

\[80-32t=0 \\ 80=32t \\ \frac{80}{32}=t \\ \frac{10}{4}=t \text{ divided top and bot by 8} \\ \frac{5}{2}=t \text{ divided top and bot by } 2\]

Answer 36

now just plug in 5/2 into the average velocity formula above

Answer 37

now i plug that back in fot t

Answer 38

well I called this t, a

Answer 39

so you replace the a in that average velocity formula I wrote above

Answer 40

yes i got it now