Question

An object is projected vertically upward from the top of a building with an initial velocity of 112 ft/sec. Its distance s(t) in feet above the ground after t seconds is given by the equation
s(t) = −16t^22 + 112t + 110.

a)Find its maximum distance above the ground.
b) Find the height of the building.

Answer 1

(a) just find the vertex of the parabola, at t = -b/2a = 112/32
(b) what is s when t=0?

Answer 2

You mean:
s(t) = −16t^2 + 112t + 110
well, for part b look at s when t = 0. It is 110 ft
for the first part a, where is the vertex of that parabola?
-s +110 = 16 t^2 -112 t completing square
t^2 - 7 t = -(1/16)( s - 110)
t^2 - 7 t + 49/4 = -1/16 (s - 110 - 49*4) = -1/16(s -306)
vertex at t = 7/2 and s = 306
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quick way with calculus
s(t) = −16t^2 + 112t + 110
ds/dt = 0 at max s = -32 t + 112
t = 112/32 = 7/2
then
s = -16(49/4) +112(7/2) + 110
= -196+ 392 + 110
= 3306 of course

Answer 3

ah, I mean 306 :)

Answer 4

Interesting. I tried these myself initially before submitting the answer here and I got the same answers as Damon. I posted here because I wasn't sure and thought I was wrong. But we ended up with the same answers.

however, the answers were wrong according to my math assignment...? If there isn't another way i'll ask my math teacher just to make sure

Answer 5

a. s(t) = -16t^2 + 112t + 110.
V = Vo + g*t = 0.
112 + (-32)t = 0,
t = 3.5 sec.

In the given Eq, replace t with 3.5 and solve for distance, s(t).
You should get 306 Ft.

b. Replace t with 0 and get 110 Ft.

Answer 6

OMG! ty so much!