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^2 + 112t + 100.
(a) Find its maximum distance above the ground.
s(t) = ft
Find the derivative of the function:
s(t) = -32t^2+112
0 = -32t^2 + 112
t=sqrt(112/32)
now once you find t plug it back into the original equation to find the max distance above the ground
usually college algebra students haven't learned about derivatives. However, s(t) is just a parabola, with its vertex at
t = 112/32
so, plug that into s(t) to find the max height
Note that you got your derivative wrong, anyway.
To find the maximum distance above the ground, we need to determine the vertex of the quadratic equation. The vertex represents the maximum or minimum point of the parabolic curve.
The equation is given by s(t) = -16t^2 + 112t + 100.
To find the vertex, we use the formula t = -b/2a, where a, b, and c are coefficients of the quadratic equation in the form ax^2 + bx + c = 0.
In this case, a = -16 and b = 112. Plugging in these values, we have:
t = -(112) / (2 * -16)
t = -112 / -32
t = 3.5
So the object reaches its maximum height after 3.5 seconds.
To find the maximum distance, substitute the value of t back into the equation:
s(t) = -16(3.5)^2 + 112(3.5) + 100
s(t) = -16(12.25) + 392 + 100
s(t) = -196 + 392 + 100
s(t) = 296
Therefore, the maximum distance above the ground is 296 feet.
To find the maximum distance above the ground, we need to determine the vertex of the quadratic equation.
The vertex form of a quadratic equation is given by the equation s(t) = a(t - h)^2 + k, where (h, k) represents the coordinates of the vertex.
In this case, the equation is already in standard form, s(t) = -16t^2 + 112t + 100. To find the vertex, we can use the formula h = -b / (2a), where a = -16 and b = 112.
h = -b / (2a)
h = -112 / (2 * -16)
h = -112 / -32
h = 3.5
Once we have the value of h, we can substitute it back into the equation to find k:
s(t) = -16t^2 + 112t + 100
s(3.5) = -16(3.5)^2 + 112(3.5) + 100
s(3.5) = -16(12.25) + 392 + 100
s(3.5) = -196 + 392 + 100
s(3.5) = 296
Therefore, the maximum distance above the ground is 296 ft.