SOLUTION: Herman arranged a game of water balloon volleyball for his daughter’s birthday party. Suppose the balloon has a height of h metres, t seconds after it is tossed, as defined by h =

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Question 1049249: Herman arranged a game of water balloon volleyball for his daughter’s birthday party. Suppose the balloon has a height of h metres, t seconds after it is tossed, as defined by h = -5t^2 + 12t + 1.
a) Write an equation that shows how many seconds will pass before the balloon hits the ground. Solve this equation,rounded to the nearest hundredth of a second.
b) what is the maximum height reached by the balloon? When does it reach this height?
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Herman arranged a game of water balloon volleyball for his daughter’s birthday party. Suppose the balloon has a height of h metres, t seconds after it is tossed, as defined by h = -5t^2 + 12t + 1.
a) Write an equation that shows how many seconds will pass before the balloon hits the ground. Solve this equation,rounded to the nearest hundredth of a second.
Height = zero when the balloon hits the ground.
Solve: -5t^2+12t+1 = 0
t = 2.48 seconds
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b) what is the maximum height reached by the balloon? When does it reach this height?
Max occurs when t = -b/(2a) = -12/(2*-5) = 6/5
Max height = f(6/5) = -5(6/5)^2+12(6/5)+1 = 8.2 metres
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Cheers,
Stan H.
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Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
(a)
+h+=+-5t%5E2+%2B+12t+%2B+1+
---------------------------
The ball is tossed at +t+=+0+
This gives you:
+h%280%29+=+-5%2A0%5E2+%2B+12%2A0+%2B+1+
+h%280%29+=+1+
They are saying that the ball is tossed from a height
of +1+ meter
---------------------------
You want to know the time in seconds when
+h+=+0+ ( balloon hits the ground )
+-5t%5E2+%2B+12t+%2B+1+=+0+
Use the quadratic formula
+t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+-5+
+b+=+12+
+c+=+1+
+t+=+%28+-12+%2B-+sqrt%28+12%5E2-4%2A%28-5%29%2A1+%29%29%2F%282%2A%28-5%29%29+
+t+=+%28+-12+%2B-+sqrt%28+144+%2B+20+%29%29%2F%28-10%29+
+t+=+%28+-12+%2B-+sqrt%28+164+%29%29%2F%28-10%29+
+t+=+%28+-12+-+12.8062+%29+%2F+%28-10%29+ ( time has to come out positive )
+t+=+24.8062+%2F+10+
+t+=+2.4806+ sec
The balloon hits the ground in 2.48 sec
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(b)
The +t+-value at the maximum is:
+t%5Bmax%5D+=+-b%2F%282a%29+
+t%5Bmax%5D+=+-12%2F%28+2%2A%28-5%29%29+
+t%5Bmax%5D+=+1.2+
Plug this value back into equation +h%28t%29+
+h+=+-5t%5E2+%2B+12t+%2B+1+
+h%281.2%29+=+-5%2A1.2%5E2+%2B+12%2A1.2+%2B+1+
+h%281.2%29+=+-5%2A1.44+%2B+14.4+%2B+1+
+h%281.2%29+=+-7.2+%2B+15.4+
+h%281.2%29+=+8.2+
The maximm height is 8.2 m which it will
reach in 1.2 sec
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check:
Here's the plot:
+graph%28+400%2C+400%2C+-4%2C+4%2C+-4%2C+10%2C+-5x%5E2+%2B+12x+%2B+1+%29+
My numbers look pretty close