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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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:
