Aball 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

c > -10.55

step-by-step explanation: