Problem Solving

An object thrown directly upward is at a height of h feet after t seconds, where h = -16(t-3)^2 + 150. At what height, in feet, is the object 2 seconds after it reaches its maximum height?

I understand that height would be at its minimum when t = 3. But how am I supposed to know when height is at its maximum???

NeilWatson wrote:An object thrown directly upward is at a height of h feet after t seconds, where h = -16(t-3)^2 + 150. At what height, in feet, is the object 2 seconds after it reaches its maximum height?

I understand that height would be at its minimum when t = 3. But how am I supposed to know when height is at its maximum???

Start with your equation, -16(t - 3)² + 150. Notice that (t - 3)² is always zero or positive, so -16(t - 3)² is always zero or negative. If it's negative, it will clearly make the height LESS than 150 feet, so the MAXIMUM value is when -16(t - 3)² is zero. It's zero at t = 3, so that's our maximum.

You're right that this sort of thinking sometimes gives you the minimum. To get technical about it, what you have here is a quadratic equation in t. Quadratic equations are of the form ax² ± bx ± c = 0. (Our equation is a quadratic too, just in t instead of in x.) If your equation equals 0, then when a is NEGATIVE, the maximum value occurs when you plug in x = -b/2a. When a is POSITIVE, as it often is in these kinds of problems, the minimum value occurs when you plug in x = -b/2a.

Taking our quadratic, we can express it as a proper quadratic as follows:

-16(t - 3)² + 150 =
-16(t² - 6t + 9) + 150 =
-16t² + 96t - 144 + 150 =
-16t² + 96t + 6

So a = -16, b = 96, c = 6. Since a is negative, the maximum occurs when we make t = -b/2a, or -96/-32, or 3.

Thank you both. When the equation was reversed it did allow me to see things more clearly. I realized this problem really isn't difficult, but under time constraints I sometimes miss these things. Still working on that.

What level problem would you consider this question?