Question

The equation $y=-16t^2+22t+45$ describes the height (in feet) of a ball thrown upwards at $22$ feet per second from $45$ feet above the ground. Find the time (in seconds) when the ball will hit the ground. Express your answer as a common fraction.

Answer 1

Set y = 0 and solve:
-16t^2 + 22t + 45 = 0
Multiply by -1: 16t^2 - 22t - 45 = 0
t = [22 ± sqrt(22^2 + 4·16·45)]/(2·16) = [22 ± 58]/32

Positive root: t = (22 + 58)/32 = 80/32 = 5/2.

So the ball hits the ground at t = 5/2 seconds.