   SEARCH HOME Math Central Quandaries & Queries  Question from mary, a student: at 9:45 Margie threw a ball upwards while standing on a platform 35ft above the ground. The height after t seconds follows the equation: h(t)= -0.6t^2 +72t+35 a) what will be the maximum height of the ball? b)how long will it take the ball reach its maximum height?? Hi Mary,

The function $h(t) = -0.6 t^2 + 72 t + 35$ is a parabola. Since the coefficient of $t^2$ is negative the parabola opens downward. I am not sure how you would determine at what value of $t$ this function reaches its maximum. If you know some calculus the differentiate $h(t)$ to obtain $h'(t)$ and then solve $h'(t) = 0$ for $t.$ If this is an algebra exercise then I expect that you know that for the parabola $y = a x^2 + b x + c$ the maximum (or minimum if $a$ is positive) occurs when $x = \large \frac{-b}{2a}.$ Whichever technique you use you will find that the maximum that $h(t)$ obtains is when $t = 60$ seconds.

The maximum height will then be $h(60)$ feet.

I really dislike this problem, it is completely ridiculous! Do you know anyone who can throw a ball upwards so that it takes a minute to reach its maximum height? Look at your value for $h(60).$ How many yards is that? How many football fields? Can you throw a ball that far?

Harley     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.