A ball is thrown into the air from the top of a building. The height, h(t), of the ball above the ground t seconds after it is thrown can be modeled by h(t) = -16t^2 + 64t + 80. How many seconds after being thrown will the ball hit the ground?

Guest Jan 28, 2020

#1**+2 **

When the ball hits the ground h will be = 0

h(t) = -16t^2 + 64t + 80

0 = h(t) = -16t^2 + 64t + 80 Solve for 't' using Quadratic Formula

One answer provided by the Quad Form will be -1 ......you want the POSITIVE answer for 't'

ElectricPavlov Jan 28, 2020