SOLUTION: At 9:45 am Sue threw a ball upwards while standing on a platform 35 feet above the ground.The height after x seconds follows the equation H(x)=-0.6X^2+72x+35. What will be the maxi
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: At 9:45 am Sue threw a ball upwards while standing on a platform 35 feet above the ground.The height after x seconds follows the equation H(x)=-0.6X^2+72x+35. What will be the maxi
Log On
Question 571884: At 9:45 am Sue threw a ball upwards while standing on a platform 35 feet above the ground.The height after x seconds follows the equation H(x)=-0.6X^2+72x+35. What will be the maximum height of the ball? And how long will it take to reach its maximum height? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! At 9:45 am Sue threw a ball upwards while standing on a platform 35 feet above the ground.The height after x seconds follows the equation H(x)=-0.6X^2+72x+35. What will be the maximum height of the ball? And how long will it take to reach its maximum height?
-----
The vertex occurs where x = -b/(2a) = -72/(-1.2) = 6 seconds
-----
Maximum height = H(6) = 445.4 ft
=====================================
Cheers,
Stan H.
============
(