is this right?
The upward line is 2s, and the total curve at the top is 8s - That seems right :)
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 =/
and we need to use derivative to find the answer
So an important general law all airborne objects follow is this: d^2 y/dt^2 = -32.2 ft/s^2
You know the second derivative of height is a constant
so that's that acceleration?
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?
The first derivative is velocity. The second derivative is acceleration We know the second derivative is constant
oh yes okay
Can you answer my question?
If youre not sure yet, that's ok too, its just kind of a shortcut if you realize it
i don't know
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?
is the first derivative v=vo-32t?
Yes, that's very well put. :)
So, do you have any ideas how we might find vo? Its ok if you dont, but if you do they could be useful
i don't really know
Dont worry, even if you cant see a way yet, we can forge ahead blindly and well find something useful eventually :P
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)
is the length in 2 sec = to the length in 8 sec?
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?
Did you follow what I said? If not, its ok.
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?
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
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
what do again in the first derivative?
You want to do it with the position equation
s = vo*t - 16*t^2
you want to set s(2) = s(10)
s=Vot-16t^2 d=Vo-32t a=-32 right?
s v and a, yes
substitute the t?
yes, you want to sub in 2 and 10 respectively for it
since you know those are equal, you can set them equal to solve for vo
and that's in units of ft/sec rememeber
Now, can you plug that into the d equation and try to use it to maximize the height?
yes thank you!!
yes thank you for your help!!
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?
6 sec? when v=0 right?
Yes! Yes! Yes! Yes!
But wait there dear
That'll be when its at the top you want to find the height then (so plug in to get s(6))
Yupee. Vo is and always will be 192 now
You just want to find s(6)
what will i do find s again? with t=6?
that will give you maximum height, and then youll be done
Yes, 576ft is the max height. You got it girl! :)
thanks again but there are still many worded problem to be solved in my book hahah
Welp. You have a great night/day. Keep working hard and don't ever give up!
i won't thank you