A ball is thrown upwards from the top of a 192-foot-tall building with an initial speed of 64 feet per second. The height of the ball as a function of time can be modeled by the function $$h(t) = –16t^2 + 64t + 192$$ What is the maximum height the ball will reach with respect to the ground?
The answer is 256 feet but I'm not sure how to get there. I tried deriving the position equation and plugging in v = 0. But I just can't get the right answer.
I am a middle school student so please do not over-complicate your answers if possible. This problem can be solved without any physics so stick to algebra/pre-calc only. Thank you!
$\endgroup$ 43 Answers
$\begingroup$First, factor your quadratic. You have$$\begin{align}h(t)&=-16t^2+64t+192\\ &=-16(t^2-4-12)\\ &=-16(t-6)(t+2)\end{align}$$
From this equation, you can see that $h(6)=0$ and $h(-2)=0$. Therefore $-2, 6$ are the zeros of this function.
The highest point of a quadratic function (if it exists) will occur at $h(x)$ where $x$ is the midpoint of the zeros. Here, $x=\frac{6-2}{2}=2$.
So, we have the maximum at $h(2)=-16(2)^2+64(2)+192=256$.
$\endgroup$ 3 $\begingroup$The height is given by a function whose graph is an upside-down parabola with respect to the $t$-axis. Compete the squares to get $$-16(t^2-4t-12)=-16(t^2-4t+4-16)=-16(t-2)^2+16^2.$$ So, you can see that there is a constant part and a part varying with time. The maximum occurs when the varying part is smallest, since it is always non positive, so that its effect is to reduce the height. This occurs when it vanishes. That is, when. $t=2.$ If you now substitute $2$ for $t$ in $h(t),$ you should get the desired maximum height reached.
On the graph, this corresponds to the vertex (turning point) of the parabola, as you would expect.
$\endgroup$ $\begingroup$The position equation is in the form of a parabola that opens downward. This is because it is in the for $f(x)=ax^2+bx+c$, where $a$ is negative. Because this parabola opens downwards, the vertex will serve as the highest point, thus it's y value will also be the highest point the ball will reach.
To find the vertex of a quadratic in this form, take $x=\frac{-b}{2a} = \frac{-64}{(2)(-16)} = 2$. This is the x coordinate of the vertex, from which you can find the y-coordinate by doing $h(2)=-16(2)^2+64(2)+192 = -64 + 128 + 192 = 256$. So the maximum height would be 256 feet.
$\endgroup$ 2
