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?
max occurs when t = -b/(2a) = -72/(2*-0.6) = -72/(-1.2) = 6 seconds
--------------------
how long will it take the ball to reach its maximum height? 6 seconds
--------------------
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)
-----
Positive solution:
t = 120.76 seconds ~ 2 min 
-------------------
the ball will hit the ground at 9:47