Back to the Ground

A ball kicked up from the ground has height $h = -5{t}^{2}+10t$ metres after t seconds. When does it land? Landing means height 0, so solve $-5{t}^{2}+10t = 0$.

There is no constant term, so both terms share the factor t, and -5 as well. Taking out -5t gives $-5t\left(t-2\right) = 0$. A product is 0 only when one of its factors is 0, so split into cases: $t = 0$ or $t-2 = 0$. The answers are 0 and 2 seconds.

Both answers are real moments. At t = 0 the ball is on the ground because it has just been kicked; at t = 2 it lands. The top of the flight is halfway between them, at t = 1, because a parabola is symmetric about its vertex.