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!
Best Answer
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 $h(2)=-16(2)^2+64(2)+192=256$.