Using calculus it can be shown that if a ball is thrown upwa

Using calculus, it can be shown that if a ball is thrown upward with an initial velocity of 16 ft/s from the top of a building 348 ft high, then its height h above the ground t seconds later will be h = 348 + 16t - 16t^2. During what time interval (in seconds) will the ball be at least 28 ft above the ground? (Enter your answer using interval notation.)

Solution

h=348 +16t-16t²

-16t²+16t+348>=28

-16t²+16t +320>=0

-16(t²-t -20)>=0

t^2 -t-20<=0

(t-5)(t+4)<=0

testing numbers t=5,-4

In [-4,5], the value is negative

And t cant be negative

Therefore the solution is [0,5]