## Calculus (3rd Edition)

The answers: (a) The particle's maximum height is 100 cm. (b) The particle hits the ground at $t=20$ s. It lands at $2040$ cm from the origin.
(a) The maximum height occurs at the critical point of $y(t)$. So, $y'(t)=\frac{d \left(20 t-t^2\right)}{\text{dt}}=20-2 t=0$ $t=\frac{20}{2}=10 s$. The particle's maximum height is $y(10)=20\times10-10^2=100\text{ cm}$. (b) The particle hits the ground when $y(t)=0$. Solving for $t$ we get, $y(t)=20 t-t^2=0$, $t(20-t)=0$, $t=0$ or $t=20$. Since the particle starts at $t=0$, so the particle hits the ground at $t=20$. Thus, it lands at $x(20)=\frac{20^3}{4}+2\times 20=2040$ cm from the origin.