SOLUTION: A ball is thrown upward from the top of a building. The ball's height above the ground after T seconds is given by the function: h(t) = -16t^2+48t+32. A. What is the initial heigh

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Question 730161: A ball is thrown upward from the top of a building. The ball's height above the ground after T seconds is given by the function: h(t) = -16t^2+48t+32.
A. What is the initial height (i.e. the height of the building)?
B. How high did the ball go?
C. When does the ball hit the ground?
Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website!
A ball is thrown upward from the top of a building. The ball's height above the ground after T seconds is given by the function: h(t) = -16t^2+48t+32.
A. What is the initial height (i.e. the height of the building)?
initial height is when t=0:
h(t) = -16t^2+48t+32
h(0) = -16(0)^2+48(0)+32
h(0) = 32 feet
.
B. How high did the ball go?
vertex is at max:
time, at vertex:
t = -b/(2a)
t = -48/(2(-16))
t = -48/(-32)
t = 3/2
.
Height at t=3/2:
h(3/2) = -16(3/2)^2+48(3/2)+32
h(3/2) = -16(9/4)+24(3)+32
h(3/2) = -4(9)+24(3)+32
h(3/2) = -36+72+32
h(3/2) = 68 feet
.
C. When does the ball hit the ground?
set h(t) to zero and solve for t:
h(t) = -16t^2+48t+32
0 = -16t^2+48t+32
0 = t^2-3t-2
solve by applying the "quadratic formula" to get:
t = {3.56, -0.56}
throw out the negative solution (extraneous) leaving
t = 3.56 seconds
.
Details of quadratic formula follows:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=17 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 3.56155281280883, -0.56155281280883. Here's your graph: