Question

A ball is thrown downward from the top of a 110-foot building with an initial velocity of 27 feet per second. The height of the ball h after t seconds is given by the equation h=-16t^2-27t+110.

How long after the ball is thrown will it strike the ground?

Answer 1

When the ball hits the ground h=0.

so, you need to solve,
0=-16t^2-25t+200

put this in the quadratic formula:

t=(25 + sqrt(225+4*16*200)) /(2*16) (usually quad formula is plus or minus but we don't care about negative time)
=4.348.