a ball is thrown up with an initial velocity of 32 ft/sec at a height of 240 ft. Use the equation h (t) = -16t2+vt+go to find when the ball hits the ground.

Step-by-step explanation:

Assuming the ball leaves the ground at a height, h = 0 ft, then the equation for the height is given by:

h = 96t - (1/2) (32) t² = 96t - 16t²

To determine the maximum height, find dh/dt, set it equal to zero, and solve for "h":

dh/dt = 96 - 32t = 0

t = 96/32 = 3 seconds (this is the time it takes for the ball to reach maximum height)

Substitute t = 3 into the original equation and solve for "h":

h (3) = 96 (3) - 16 (3) ² = 288 - 144 = 144 feet