A ball is thrown downward from the top of a 180 foot building with an initial velocity of 20 feet per second. The height of the ball h after t seconds is given by the equation h=-16t^2-20t+180. How long after the ball is thrown will it be 50 feet from the ground? Round the result to the nearest tenth of a second.

Solution:

h=-16t^2-20t+180 = 50

-16t^2 -20t +130 = 0

t= 2.29

So it takes 2.29 seconds to be 50 feet from the ground