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???
Formula Problem
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- NeilWatson
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The formula h = -16 (t - 3)² + 150 allows us to determine the height of the object at any time. For what value of t is -16(t-3)² + 150 maximized (in other words, the object is at its maximum height)?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?
a) 6
b) 86
c) 134
d) 150
e) 166
It might be easier to answer this question if we rewrite the formula as h = 150 - 16(t-3)²
To MAXIMIZE the value of h, we need to MINIMIZE the value of 16(t-3)² and this means minimizing the value of (t-3)²
As you can see,(t-3)² is minimized when t = 3.
We want to know the height 2 seconds AFTER the object's height is maximized, so we want to know that height at 5 seconds (3+2)
At t = 5, the height = 150 - 16(5 - 3)²
= 150 - 16(2)²
= 150 - 64
= 86
Answer: B
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Brent
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The MAXIMUM height occurs at t = 3.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???
To verify this, try plugging in some different values of t
Cheers,
Brent
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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.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???
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.
- NeilWatson
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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.
- NeilWatson
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I'd say 500-600 levelNeilWatson wrote:What level problem would you consider this question?
Cheers,
Brent