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)?
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:
*[invoke quadratic "x", 1, -3, -2 ]