Aball is thrown from a height of 80 feet with an initial downward velocity of 4/fts . the ball's height h (in feet) after t seconds is given by the following.

h−80−4t-16t^2

how long after the ball is thrown does it hit the ground? round your answer(s) to the nearest hundredth.

Alright, so we have h=80-4t-16t^2. We have to find when h=0. We can use the quadratic formula

(use x as t)

x=(4+-sqrt(16-(-64*80))/32=(4+-sqrt(5136))/(-32) , and since 4+sqrt(5136)/(-32) is clearly negative that doesn't work, so we have (4-sqrt(5136))/(-32), which is approximately 2.11

a

step-by-step explanation:

a

29933333.33

step-by-step explanation: