<PRE><B>Question 1049249</B>
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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