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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?

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    $\begingroup$ $t^2-10t+24=(t-6)(t-4)$. $\endgroup$ – Jared May 17 '13 at 4:21
  • $\begingroup$ I'm an idiot.. I was concentrating on 12 and 2, thanks guys $\endgroup$ – Seb May 17 '13 at 4:23
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    $\begingroup$ Not a problem, @Foxic!! ;-) $24$ has a lot of factors: $1, 2, 3, 4, 6, 8, 12, 24$ $\endgroup$ – Namaste May 17 '13 at 4:28
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$$16(t^2 - 10 t + 24) = 0 \iff 16(t-4)(t-6) = 0 \implies t = 4,\;\; t = 6$$

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 $$\;t^2 - 10 t + 24 = (t - 4)(t - 6)$$

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  • $\begingroup$ With $s(t) = 160 t - 16 t^2$, $v = \frac {ds}{dt} = 160 - 32 t$. Now that you have your $t$-values, you can solve for velocity, $v(4), v(6)$ $\endgroup$ – Namaste May 17 '13 at 4:33
  • $\begingroup$ Always helpful and descriptive +1 $\endgroup$ – Amzoti May 17 '13 at 5:19
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$$\text{So, }v=\frac{ds}{dt}=160-32t$$

$$\text{Now, }160t-16t^2=384\implies t^2-10t+24=0\implies t=4,6$$

The smaller value of $t$ will occur during upward movement

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If we know that the formula for motion under uniform acceleration is $s = v_0 t + \frac12 at^2,$ where $v_0$ is the initial velocity and $a$ is the acceleration, then a glance at the formula $s = 160 t - 16t^2$ tells us that $v_0=160$ (which was given in the problem statement anyway) and $a = -32.$

It therefore takes $(-160)/(-32) = 5$ seconds to reduce the upward velocity from $160$ to $0.$

At the end of $5$ seconds, the ball's height is $160(5)-16(5^2) = 400$ feet and it is at the highest point it will reach. It follows that the plot of the ball's height over time is a parabola with vertex at $t=5$ and formula $s = 400 - 16(t-5)^2.$

We plug in $s = 384,$ that is, $384 = 400 - 16(t-5)^2.$ Subtracting $400,$ we have $-16 = - 16(t-5)^2,$ so we just need to divide by $-16$ and then take square roots to determine that $t - 5 = \pm 1.$

So its velocity on the way down is just the velocity an object has after falling $1$ second from a rest position (which is a negative value), and to get the velocity on the way up we just reverse the sign. Or we can do it "the hard way" and compute the velocities at $t=5-1$ and $t = 5 + 1.$


If we know about potential and kinetic energy, however, then it's just plug-and-chug. The kinetic energy at ground level is $\frac12mv_0^2 = \frac12m(160^2) = 12800m,$ where $m$ is the mass of the ball. Setting potential at the ground level to zero, the total energy is therefore $12800m.$ The potential energy at $384$ feet is $mgs = m\times32\times384 = 12288m.$ Since kinetic plus potential is constant, that means $12288m + KE_{384} = 12800m,$ where $KE_{384}$ is the kinetic energy at $384$ feet. So $KE_{384} = 12800m - 12288m = 512m = \frac12mv_{384}^2,$ where $v_{384}$ is the velocity at $384$ feet; so $v_{384}^2 = 1024.$ Taking square roots, $v_{384} = \pm 32.$

Lots of things get simpler when we know how to apply conservation of energy.

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