I believe this is a related rates problem.
I attempted part 2 but I'm not sure about the equation for the maximum height. is it the derivative ?
To clarify what you wrote,part A.
it is a parabola with a max , so the maximum height will be at the max point in the graph.
s(t) = 48t+64t-16t^2
take derviative . s'(t) = 64 -32t
set to zero t= 2 . plug t back in original?
48(2)+64(2)-16(2)^2 = 160
You started with an equation: 48 + 64t - 16t^2 = 0.Part B.
i did what you suggested.
s(t) = 48+64t-16t^2 = 0
I get -(t+1) (t+4) sooo t= -1 and t=-4 ??
i know what to do next but im getting two "t" values right now ..