#### Answer

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.

#### Work Step by Step

(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.