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anonymous
 one year ago
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.
anonymous
 one year ago
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.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443848025630:dw

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3The upward line is 2s, and the total curve at the top is 8s  That seems right :)

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3We 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 =/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and we need to use derivative to find the answer

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3So an important general law all airborne objects follow is this: d^2 y/dt^2 = 32.2 ft/s^2

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3You know the second derivative of height is a constant

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so that's that acceleration?

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3Lets 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
 one year ago
Best ResponseYou've already chosen the best response.3The first derivative is velocity. The second derivative is acceleration We know the second derivative is constant

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3Can you answer my question?

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3If youre not sure yet, that's ok too, its just kind of a shortcut if you realize it

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3That'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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is the first derivative v=vo32t?

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3Yes, that's very well put. :)

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3So, do you have any ideas how we might find vo? Its ok if you dont, but if you do they could be useful

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3Dont worry, even if you cant see a way yet, we can forge ahead blindly and well find something useful eventually :P

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3So 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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is the length in 2 sec = to the length in 8 sec?

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3@jarp0120 you said before: s= Vot16t^2? That's technically correct...did you have a d0 at first, and replace it with zero? or an s0, I guess

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443853241921:dw

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3So 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
 one year ago
Best ResponseYou've already chosen the best response.3Did you follow what I said? If not, its ok.

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3Well 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?

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3So 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

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3It 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what do again in the first derivative?

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3You want to do it with the position equation

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3you want to set s(2) = s(10)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0s=Vot16t^2 d=Vo32t a=32 right?

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3v = vo  32t, I think you meant? @jarp0120

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3yes, you want to sub in 2 and 10 respectively for it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0s(2)=2Vo64 s(10)=10Vo1600

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3since you know those are equal, you can set them equal to solve for vo

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3and that's in units of ft/sec rememeber

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3Now, can you plug that into the d equation and try to use it to maximize the height?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes thank you for your help!!

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3No 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.06 sec? when v=0 right?

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3That'll be when its at the top you want to find the height then (so plug in to get s(6))

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3Yupee. Vo is and always will be 192 now

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3You just want to find s(6)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what will i do find s again? with t=6?

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3that will give you maximum height, and then youll be done

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3Yes, 576ft is the max height. You got it girl! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks again but there are still many worded problem to be solved in my book hahah

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3Hahaha. #thestruggleisreal

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0#calculusislove hahahah

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.3Welp. You have a great night/day. Keep working hard and don't ever give up!
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