If not possible, enter IMPOSSIBLE.)When does the ball initially reach a height of 64 ft?

When does the ball hit the ground

Since the ball feels the effects of gravity, use Constant Acceleration Equations:HEIGHT of 118x=vi*t+1/2*at^2118 = 80t + 1/2*-9.8*t^2118 = 80t - 4.9t^24.9t^2 - 80t + 118 = 0Use quadratic formula to solve for t

GREATEST HEIGHT/HIGHEST POINT OF PATH, final velocity (vf = 0), acceleration is gravity (a = -9.8)vf^2=vi^2+2ax0=80^2+2(-9.8)x19.6x=6400x = 326.5

HITS GROUND at twice the time it takes to reach highest pointvf=vi+atAt highest point, vf=00 = 80 + -9.8*t9.8t = 80t = 8.162t = 16.32

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