Question 831437
The ball hits the ground when {{{h=0}}} so we solve
{{{-16t^2+64t+264=0}}}
Dividing both sides of the equation by 8, it becomes
{{{-2t^2+8t+33=0}}}
Using the quadratic formula we find two solutions given by
{{{t = (-8 +- sqrt(8^2-4*(-2)*33 ))/(2*(-2)) }}}
{{{t = (-8 +- sqrt(64+264))/(-4) }}}
{{{t = (-8 +- sqrt(328))/(-4) }}}
{{{t = (8 +- sqrt(328))/4 }}} .
We take {{{t=(8+sqrt(328))/4=(4+sqrt(82))/2}}} because the other solution is negative.
The approximate value is {{{highlight(6.53seconds)}}} .
 
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.