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.
(a) just find the vertex of the parabola, at t = -b/2a = 112/32
(b) what is s when t=0?
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
ah, I mean 306 :)
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
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.
OMG! ty so much!
To find the maximum distance above the ground, we need to find the vertex of the quadratic function s(t) = -16t^2 + 112t + 110. The vertex represents the maximum point of the parabola.
The formula for the x-coordinate of the vertex of a quadratic function in the form of ax^2 + bx + c is given by x = -b / (2a).
In this case, the coefficient of t^2 is a = -16, and the coefficient of t is b = 112. Let's substitute these values into the formula:
x = -112 / (2 * (-16))
x = -112 / (-32)
x = 3.5
So, the time at which the object reaches its maximum distance above the ground is 3.5 seconds. Now, let's substitute this value of t back into the equation to find the maximum distance:
s(3.5) = -16(3.5)^2 + 112(3.5) + 110
s(3.5) = -16(12.25) + 392 + 110
s(3.5) = -196 + 392 + 110
s(3.5) = 306
Therefore, the maximum distance above the ground is 306 feet.
To find the height of the building, we need to find the value of s(t) when t = 0. This represents the initial height of the object above the ground.
s(0) = -16(0)^2 + 112(0) + 110
s(0) = 0 + 0 + 110
s(0) = 110
So, the initial height of the object above the ground is 110 feet. Therefore, the height of the building is 110 feet.