A flying cannonball’s height is described by formula $y=−16t^2+300t$. Find the highest point of its trajectory. In how many seconds after the shot will cannonball be at the highest point? What is the highest point?
I tried finding the vertex but it does not seem to help me. I know how to graph it, but I just don't know how to use it. Thanks!
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If you can't use derivatives, you may recall from your class that the value of the $t$-coordinate of the vertex of the equation $y = at^2+bt+t$ is $\frac{-b}{2a}$ (using calculus, you simply just derive that expression rather than take it for granted). So in this problem, the $x$-coordinate of the vertex is $\frac{-300}{-32} = 9.375$
Plugging that in you get $y = -16(9.375)^2 + 300(9.375) = 1406.25$.
The $t$ coordinate of your vertex answers the first question, and the $y$ coordinate answers the second question. I assume your work looked like this:
$$y=-16t^2+300t$$
$$y'=-32t+300$$
$$0=-32t+300$$
$$t=9.375^{*}$$
$$y=-16(9.375)^2+300(9.375)$$
$$y=1406.25^{**}$$
*Answer to first question
**Answer to second question