If a ball is thrown vertically upward with a velocity of $160 \text{ ft/s}$, then its height after t seconds is $s = 160t − 16t^2$.
a) What is the velocity of the ball when it is $384 \text{ ft}$ above the ground on its way up?
b) What is the velocity of the ball when it is $384 \text{ ft}$ above the ground on its way down?
$$\begin{align*}
384 &= 160t – 16t^2\\
16t^2 – 160t + 384 &= 0\\
16 (t^2 – 10t + 24) &= 0
\end{align*}$$
I can't factor the above… and I'm not supposed to used the quadratic formula. Am I stupid or is this unsolvable without without the formula?
Best Answer
$$16(t^2 - 10 t + 24) = 0 \iff 16(t-4)(t-6) = 0 \implies t = 4,\;\; t = 6$$
Heading up: $384$ feet at time $t = 4$,
Descendng down after reaching maximum height: $384$ ft. at time $t = 6$.
Note: $$(-4)\cdot (-6) = + 24;\quad -4 + - 6 = -10$$
Noticing those facts allow you to deduce that the factors must be $$\;t^2 - 10 t + 24 = (t - 4)(t - 6)$$