The height h in feet of a ball thrown into the air after t seconds is given by h(t)= -16t squared +32t+3. Find the height of the ball after 2.5 seconds

Answer 1

h(t) = -16t^2 + 32t + 3

h(2.5) = -16(2.5)^2 + 32(2.5) + 3

= -100+ 80 + 3 = - 17 feet below ground

max height is h'(t) = -32t + 32 = 0 at time t=1 h(1) = -16+32 + 3 = 19 feet

it comes back to ground level when h(t)=0 = -16t^2 + 32t + 3 = 0

t < 2.5