An object is launched at 20 m/s from a height of 65 m. The equation for the height (h) in terms of time (t) is given by h(t) = -4.9t2 + 20t + 65. What is the object's maximum height?

Question

anonymous · Answer

I'm guessing that we want to find the vertex of the function....not exactly sure...what do you think?

lgbasallote · Answer

it would be when final velocity is zero

anonymous · Answer

Final velocity is zero when it hits the ground.

anonymous · Answer

Yeah I believe vertex is what we are trying to find, im not quite sure

lgbasallote · Answer

what i would do is a little complicated lol...idk how to do this in simple way...but i would take the derivative then equate to 0..solve for t...then sub that value of t to h(t)

lgbasallote · Answer

i dont think you got to derivative yet..

anonymous · Answer

That's what I was thinking....

lgbasallote · Answer

but for the sake of it i'd like to do it :DDD

h'(t) = -9.8 t + 20

0 = -9.8t + 20
9.8t = 20
t = 2.04

now sub back in to original equation

h(t) = -4.9 (2.04)^2 + 20(2.04) + 65
h(t) = -20.4 + 40.8 + 65
h(t) = 85.4

lgbasallote · Answer

idk how to do it simply though

lgbasallote · Answer

finding the maximum using algebra is complicated lol

anonymous · Answer

thats okay thanks much! :D

anonymous · Answer

Please don't give answers

anonymous · Answer

@Romero, if @iceywheniplease7 wanted answers, she'll get it.