グローバルビジネスで役立つ数学（18）放物線上の高さと時間（英語版）

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
【レッスン】私のオンライン英語レッスンをご希望の方はこちらをご覧ください。
【コンテント】当サイトで提供する情報はその正確性と最新性の確保に努めていま
　すが完全さを保証するものではありません。当サイトの内容に関するいかなる間
　違いについても一切の責任を負うものではありません。