Question

a ball is thrown into the air with an upward velocity of 36 ft/s. it’s height h in feet after t seconds is given by the function h=-16t^2+36t+9. in how many seconds does the ball reach its maximum height?

Answer 1

Complete the square:

`h(t)=-16t^2+36t+9=-16\left(t^2-\frac94t\right)+9=-16\left(\left(t-\frac98\right)^2-(81)/(64)\right)+9`

`h(t)=-16\left(t-\frac98\right)^2+\frac{117}4`

That is, the graph of
`h(t)` is a parabola with vertex at
`(t,h(t))=\left(\frac98,\frac{117}4\right)`, meaning the ball reaches its maximum height of
`\frac{117}4=29.25\,\mathrm{ft}` after
`t=\frac98=1.125\,\mathrm s` of being thrown.