Hi, there!
The word problem today requires calculation abilities in that mathematics
develops into physics. Mathematics is already beyond my capacity; how-
ever, it is my disposition to be driven by an impulse to challenge a physics
word problem standing in my way.
An object is launched from ground level directly upward at \(39.2\,m/s\).
For how long is the object at or above \(58.8\) meters high?
First of all, let’s take a look at the following formula.
\(h = v_{0}t – \frac{1}{2}gt^{2}\)
\(h\) = height
\(v_{0}\) = initial velocity
\(g\) = acceleration of gravity
\(t\) = time
The object started at ground level; thus initial height was \(0\). Since the ac-
celeration of gravity is \(9.80665\,m/s^{2}\), the gravity number will be “\(4.9\)”.
The equation is:
\(h(t) = -4.9t^{2} + 39.2t\)
This is a negative quadratic, so the graph is an upside-down parabola.
We can find twice when the object is precisely \(58.8\) meters high. So let’s
solve the following:
\(–4.9t^{2} + 39.2t = 58.8\)
\(4.9t^{2} – 39.2t + 58.8 = 0\)
\(t^{2} – 8t + 12 = 0\)
\((t – 6)(t -2) =0\)
Then the object is at \(58.8\) meters two seconds after launch and six sec-
onds after launch coming back down. Subtracting to find the difference,
we find that the object is at or above \(58.8\) meters for four seconds.
As I get older, I seldom throw an object horizontally or vertically. I am a
bit apprehensive about the time I won’t be able to pick even an eraser in
my hand.
Stay tuned, and expect to see my next post.
Keep well.
Frank Yoshida
￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣
【グローバルビジネスで役立つ数学】でもっと学習する b＾＾)
【参考図書】『もう一度高校数学』（著者：高橋一雄氏）株式会社日本実業出版社
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