SOLUTION: at 9:45 am Maggie threw a ball upwards while standing on a platform 55 ft off of the ground. The trajectory after t seconds follows the equation: h(t)=-0.6t^2 + 72t + 55 what will

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Question 461487: at 9:45 am Maggie threw a ball upwards while standing on a platform 55 ft off of the ground. The trajectory after t seconds follows the equation: h(t)=-0.6t^2 + 72t + 55
what will be the maximum height of the ball
how long will it take the ball to reach its maximum height
at what time will the ball hit the ground
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
at 9:45 am Maggie threw a ball upwards while standing on a platform 55 ft off of the ground. The trajectory after t seconds follows the equation:
h(t) = -0.6t^2 + 72t + 55
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What will be the maximum height of the ball?
max occurs when t = -b/(2a) = -72/(2*-0.6) = -72/(-1.2) = 6 seconds
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how long will it take the ball to reach its maximum height? 6 seconds
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at what time will the ball hit the ground
Solve -0.6t^2+72t+55 = 0
Use the quadratic formula:
t = [-72 +- sqrt(72^2 - 4*-0.6*55)]/(2*-0.6)
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Positive solution:
t = 120.76 seconds ~ 2 min
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the ball will hit the ground at 9:47