a ball is thrown vertically upward with an initial velocity of 160 ft per second The distance in feet of the ball from ground after t seconds is s=160t-16t^2

Question

for what interval of time is the ballmore than 256 above ground

Answer 1

-16t^2 + 160t > 256

-16t^2 + 160t - 256 > 0

16t^2 - 160t + 256< 0

f(t) = t^2 - 10t + 16 < 0

(t - 8)(t - 2 ) < 0

for x<2, f(x) > 0

for 2<x < 8, f(x)<0

for x>8, f(x)>0

so the ball is higher than 256 for 2<t<8