Maximum height of a projectile

Text

A body thrown vertically upward has height $h(t)=30t-5t^{2}$ (metres, seconds). Find the instant at which it reaches maximum height and that height.

Solution

$h'(t)=30-10t=0\Rightarrow t=3\text{ s}$. Since $h''(t)=-10<0$ it is a maximum. $h(3)=90-45=45\text{ m}$. $\boxed{t=3\ \text{s};\quad h_{\max}=45\ \text{m}}$