a ball is thrown upward with an initial velocity of 32 ft/s from a height 768 ft.

h= 016t^2+32t+768

after how many seconds will the ball reach the ground?

I.e. for what positive value of t is h=0?

Is there 0.16 or just 16?

The answer will be in a few moments.

-16...sorry

-16t^2 + 32t + 786 = 0

D= 32^2 - 4*(-16)*786 = 1024 + 50304 = 51328 = (8sqrt(802))^2

t1 = (-32+226.56)/2*(-16)= -6 (is not a solution because time cannot be negative)

t2 = (-32-226.56)/2*(-16) = 8

answer is 8 seconds

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