- jarp0120

calculus:rectilinear motion
an object is thrown vertically upward from the ground reaches a certain height after 2 sec and returns to the same height on descent, 8 sec later. find its initial velocity and the height in question.

- chestercat

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- jarp0120

|dw:1443848025630:dw|

- jarp0120

is this right?

- Miracrown

The upward line is 2s, and the total curve at the top is 8s - That seems right :)

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## More answers

- Miracrown

We want to use the idea that acceleration due to gravity is constant to help us
Do you know if your class prefers feet or meters? We can use either units, but I don't want to do it in a way you haven't seen =/

- Miracrown

- jarp0120

feet

- jarp0120

and we need to use derivative to find the answer

- Miracrown

So an important general law all airborne objects follow is this:
d^2 y/dt^2 = -32.2 ft/s^2

- Miracrown

You know the second derivative of height is a constant

- jarp0120

so that's that acceleration?

- Miracrown

Lets think mostly about the 8 seconds it was highest up first. What do you think the first derivative of its height was when it was at the very top?

- Miracrown

The first derivative is velocity.
The second derivative is acceleration
We know the second derivative is constant

- jarp0120

oh yes okay

- Miracrown

Can you answer my question?

- Miracrown

If youre not sure yet, that's ok too, its just kind of a shortcut if you realize it

- jarp0120

i don't know

- Miracrown

That's ok, lets move forward with the problem a bit and readdress it later, when it comes up. So we have this constant second derivative...But that's not very descriptive.
Can you use it to get an equation for the first derivative?

- jarp0120

is the first derivative v=vo-32t?

- Miracrown

Yes, that's very well put. :)

- Miracrown

So, do you have any ideas how we might find vo?
Its ok if you dont, but if you do they could be useful

- jarp0120

i don't really know

- Miracrown

Dont worry, even if you cant see a way yet, we can forge ahead blindly and well find something useful eventually :P

- jarp0120

thank you

- Miracrown

So often when unsure the best thing to do in real life is just fight forward. So lets do that.
We found an equation for velocity. Can you find one for height? (y, or the 0th derivative)

- jarp0120

s=vo-16t^2?

- jarp0120

s= Vot-16t^2?

- jarp0120

- jarp0120

is the length in 2 sec = to the length in 8 sec?

- Miracrown

@jarp0120 you said before: s= Vot-16t^2?
That's technically correct...did you have a d0 at first, and replace it with zero?
or an s0, I guess

- jarp0120

|dw:1443853241921:dw|

- Miracrown

So we have an equation for s. But we still dont know vo.Maybe we can figure it out though
What two times is the ball at the same height?

- Miracrown

- jarp0120

what?

- Miracrown

Did you follow what I said? If not, its ok.

- jarp0120

no hahaha

- Miracrown

Well remember how the problem is framed, It takes 2s to go up to a certain point
and then 8 more seconds to go up, and back down to that point.
So wouldnt it have the same height at t=2s and t=10s?

- jarp0120

yeah

- Miracrown

So how about you plug in 2 and 10 into your position equations and set them equal
since we know 2 and 10 should give the same result

- jarp0120

yes sir

- Miracrown

It was a little bit of a trick there. So, try out my setting equal idea
you should be able to use it to solve vo

- jarp0120

what do again in the first derivative?

- Miracrown

You want to do it with the position equation

- Miracrown

s = vo*t - 16*t^2

- Miracrown

you want to set s(2) = s(10)

- jarp0120

s=Vot-16t^2
d=Vo-32t
a=-32
right?

- Miracrown

s v and a, yes

- Miracrown

v = vo - 32t, I think you meant? @jarp0120

- jarp0120

substitute the t?

- jarp0120

yes

- Miracrown

yes, you want to sub in 2 and 10 respectively for it

- jarp0120

s(2)=2Vo-64
s(10)=10Vo-1600

- Miracrown

Mhm

- Miracrown

since you know those are equal, you can set them equal to solve for vo

- jarp0120

yay!!!!
Vo=192

- Miracrown

Hooray! lol

- Miracrown

and that's in units of ft/sec rememeber

- Miracrown

Now, can you plug that into the d equation and try to use it to maximize the height?

- jarp0120

yes thank you!!

- Miracrown

yw :D

- jarp0120

salamat po!!

- jarp0120

yes thank you for your help!!

- Miracrown

No worries. But hand on, It wants you to find the maximum height too doesn't?
Alternately, you can maybe guess when the height is maxed
If its the same height at t=2 and t=10
When do you think its at the top of its arc?

- jarp0120

wait

- Miracrown

tyt

- jarp0120

6 sec?
when v=0 right?

- jarp0120

:)

- Miracrown

Yes! Yes! Yes! Yes!

- Miracrown

But wait there dear

- Miracrown

That'll be when its at the top you want to find the height then
(so plug in to get s(6))

- jarp0120

when Vo=192?

- Miracrown

Yupee. Vo is and always will be 192 now

- jarp0120

and then

- Miracrown

You just want to find s(6)

- jarp0120

what will i do find s again? with t=6?

- Miracrown

Yes

- jarp0120

576 ft

- Miracrown

that will give you maximum height, and then youll be done

- Miracrown

Yes, 576ft is the max height. You got it girl! :)

- jarp0120

thanks again
but there are still many worded problem to be solved in my book hahah

- Miracrown

Hahaha. #thestruggleisreal

- jarp0120

#calculusislove hahahah

- Miracrown

Welp. You have a great night/day. Keep working hard and don't ever give up!

- jarp0120

i won't thank you

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