SOLUTION: The height h (in feet) above the ground of a baseball depends upon the time t (in seconds) it has been in flight. Cameron takes a mighty swing, but hits a bloop sinlge whose height

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Question 133135: The height h (in feet) above the ground of a baseball depends upon the time t (in seconds) it has been in flight. Cameron takes a mighty swing, but hits a bloop sinlge whose height is described approximately by the equation h=80t-16t^2. Ho long is the ball in the air? The ball reaches its maximum heght after how many seconds of light? What is the maximum height? How much time does it take for the ball to reach a hieght of 60 feet? On its way back down, the ball is again 60 feet above the ground; what is the value of t when this happens?
Found 2 solutions by ankor@dixie-net.com, stanbon:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The height h (in feet) above the ground of a baseball depends upon the time t (in seconds) it has been in flight. Cameron takes a mighty swing, but hits a bloop single whose height is described approximately by the equation h=80t-16t^2. How long is the ball in the air? The ball reaches its maximum height after how many seconds of flight? What is the maximum height? How much time does it take for the ball to reach a height of 60 feet? On its way back down, the ball is again 60 feet above the ground; what is the value of t when this happens?
:
Find how long the ball will be in the air by solving for h = 0
-16t^2 + 80t = 0
Factor out -t:
-t(16t - 80) = 0
Two solutions:
-t = 0
and
16t = +80
t = 80/16
t = 5 secs for the ball to hit the ground
:
You can find the time of max height by finding the axis of symmetry:
Using the form: y = ax^2 + bx + c; the axis of symmetry: x = -b/(2a)
:
Putting the equation in this form, we have: h = -16t^2 + 80t: a=-16, b = 80
t = -80/(2*-16)
t = -80/-32
t = +2.5 seconds to reach max height
:
We find the max height by substituting 2.5 for t in the original equation
h = 80(2.5) - 16(2.5^2)
h = 200 - 16(6.25)
h = 200 - 100
h = 100 ft is the max height
:
We find the time for 60ft of height by substituting 60 for h in the original equation
60 = 80t - 16t^2
arrange a quadratic equation on the left
16t^2 - 80t + 60 = 0
Use the quadratic formula to solve for t:
t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
:
t+=+%28-%28-80%29+%2B-+sqrt%2880%5E2+-+4+%2A+16+%2A+60+%29%29%2F%282%2A16%29+
:
t+=+%2880+%2B-+sqrt%286400+-+3840+%29%29%2F%2832%29+
:
Solve this and you will have two solutions of:
t = .92 sec, 60 ft on the way up
t = 4.08 sec, 60 ft on the way down
:
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Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
height is described approximately by the equation h=80t-16t^2.
How long is the ball in the air?
Height = zero when the ball is back on the ground:
-16t^2+80t = 0
-16t(t-5) = 0
t = 0 sec or t = 5 sec.
The ball is in the air 5 seconds.
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The ball reaches its maximum height after how many seconds of flight?
What is the maximum height?
max = -b/2a = -80/(-32) = 5/2 seconds
max height is h(5/2) = 80(5/2)-16(5/2)^2 = 200 -100 = 100 ft.
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How much time does it take for the ball to reach a height of 60 feet?
-16t^2+80t = 60
Divide thru by -4 to get:
4t^2-20t+15 = 0
t = [20 +- sqrt(400-4*4*15)]/8
t = [20 +- sqrt(160)]/8
Positive solution:
or t = [20-4sqrt(10)]/8
t = 0.918861 seconds
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On its way back down, the ball is again 60 feet above the ground; what is the value of t when this happens?
t = [20 + 4sqrt(10)]/8
t = 4.0811 seconds
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Cheers,
Stan H.