SOLUTION: A person standing close to the edge on the top of a 264 foot building throws a baseball vertically upward. The quadratic equation h=-16t^2+64t+264 models the ball's height

Algebra ->  Finance -> SOLUTION: A person standing close to the edge on the top of a 264 foot building throws a baseball vertically upward. The quadratic equation h=-16t^2+64t+264 models the ball's height       Log On


   

Question 831437: A person standing close to the edge on the top of a 264 foot building throws a baseball vertically upward. The quadratic equation
h=-16t^2+64t+264

models the ball's height above the ground in feet, seconds after it was thrown.
How high is the ball after 2 seconds?


How many seconds does it take until the ball finally hits the ground? Round to the nearest tenth of a second.

I got the first part but can't figure out the second.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The ball hits the ground when h=0 so we solve
-16t%5E2%2B64t%2B264=0
Dividing both sides of the equation by 8, it becomes
-2t%5E2%2B8t%2B33=0
Using the quadratic formula we find two solutions given by
t+=+%28-8+%2B-+sqrt%288%5E2-4%2A%28-2%29%2A33+%29%29%2F%282%2A%28-2%29%29+
t+=+%28-8+%2B-+sqrt%2864%2B264%29%29%2F%28-4%29+
t+=+%28-8+%2B-+sqrt%28328%29%29%2F%28-4%29+
t+=+%288+%2B-+sqrt%28328%29%29%2F4+ .
We take t=%288%2Bsqrt%28328%29%29%2F4=%284%2Bsqrt%2882%29%29%2F2 because the other solution is negative.
The approximate value is highlight%286.53seconds%29 .

NOTES:
Physics says that the person threw the ball up at an initial speed of
64 ft/sec.
Physics also says that the wording to the problem is wrong, because the equation says that the ball starts at 264 feet over the ground. That would be where the hand of the person standing close to the edge on the top of the building was as the ball left the hand. I would expect that hand to be 3 to 4 feet above the level the person is standing on. The problem did not need to mention the height of the building.