A ball is thrown upward from the top of a tower.If its height is described by $-t ^2+60t+700$,what is the greatest height the ball attains?

Question

A ball is thrown upward from the top of a tower.If its height is described by $-t ^2+60t+700$,what is the greatest height the ball attains ?

I've managed to get a solution by realizing that the time at which the ball attains the maximum must be the midpoint between the time its height is $0$,so: $$-(t^2-60t-700)=0$$ $$-(t-70)(t+10)=0$$

So I have that when $t=70$ and $t=-10$ ,$h(t)=0$ ,now the midpoint is $\cfrac{70-(-10)}{2}-10 =30$ since the parabola is shifted $-10$ to the origin.So the maximum is attained at $t_{m}=30$ at that point $h(t)=1600$.

However this is not the solution intented for the problem as my book doesn't teach any conics (this problem actually comes from the chapter about inequalities),but I don't see any other way to solve this.

What is the easier solution intended to the problem ?

Answer 1

Complete the square:

$$-t^2+600t+700 = -(t^2 - 60t + 900 - 900 - 700) = -((t^2-30)^2-1600).$$

So it's an upside down parabola with vertex $(30, 1600)$. The maximal value is thus $h = 1600$ as you have obtained.

Answer 2

HINT:

$$-t^2+60t+700=900-(t-30)^2+700\le900+700$$ for real $t$