A ball is dropped from the top of a 256-foot-high building. The ball is 192 feet

Trent Carpenter 2021-10-17 Answered
A ball is dropped from the top of a 256-foot-high building. The ball is 192 feet above ground after 2 seconds, and it reaches ground level in 4 seconds. The height above ground is a quadratic function of the time after the ball is thrown. Write the equation of this function.

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opsadnojD
Answered 2021-10-18 Author has 18503 answers

A ball is dropped from the top of a 256-foot-high building. The ball is 192 feet above ground after 2 seconds, and it reaches ground level in 4 seconds.
Let the height above the ground after t seconds be,
\(\displaystyle{h}{\left({t}\right)}={a}{t}^{{{2}}}+{b}{t}+{c}\) \(\rightarrow\) (1)
Since the initial height is 256 foot, therefore \(\displaystyle{h}{\left({t}\right)}={256}\) when \(\displaystyle{t}={0}\).
Now put \(\displaystyle{t}={0}\) and \(\displaystyle{h}{\left({t}\right)}={256}\) in equation (1), we get
\(\displaystyle\Rightarrow{256}={a}{\left({0}\right)}^{{{2}}}+{b}{\left({0}\right)}+{c}\)
\(\displaystyle\Rightarrow{c}={256}\)
Since the ball is 192 feet above ground after 2 seconds, therefore \(\displaystyle{h}{\left({t}\right)}={192}\) when \(\displaystyle{t}={2}\).
Now put \(\displaystyle{t}={2}\) and \(\displaystyle{h}{\left({t}\right)}={256}\) in equation (1), we get
\(\displaystyle\Rightarrow{192}={a}{\left({2}\right)}^{{{2}}}+{b}{\left({2}\right)}+{c}\)
\(\displaystyle\Rightarrow{4}{a}+{2}{b}+{c}={192}\)
\(\displaystyle\Rightarrow{4}{a}+{2}{b}+{256}={192}\)
\(\displaystyle\Rightarrow{4}{a}+{2}{b}=-{64}\)
\(\displaystyle\Rightarrow{2}{a}+{b}=-{32}\) \(\displaystyle\rightarrow\) (2)
Since the ball reaches the ground level after 4 seconds, therefore \(\displaystyle{h}{\left({t}\right)}={0}\) when \(\displaystyle{t}={4}\).
Now put \(\displaystyle{t}={4}\) and \(\displaystyle{h}{\left({t}\right)}={0}\) in equation (1), we get
\(\displaystyle\Rightarrow{0}={a}{\left({4}\right)}^{{{2}}}+{b}{\left({4}\right)}+{c}\)
\(\displaystyle\Rightarrow{16}{a}+{4}{b}+{c}={0}\)
\(\displaystyle\Rightarrow{16}{a}+{4}{b}+{256}={0}\)
\(\displaystyle\Rightarrow{16}{a}+{4}{b}=-{256}\)
\(\displaystyle\Rightarrow{4}{a}+{b}=-{64}\) \(\displaystyle\rightarrow\) (3)
Now subtracting equation (2) from equation (3), we get
\(\displaystyle{\left({4}{a}+{b}\right)}–{\left({2}{a}+{b}\right)}=-{64}–{\left(-{32}\right)}\)
\(\displaystyle\Rightarrow{2}{a}=-{32}\)
\(\displaystyle\Rightarrow{a}=-{16}\)
Now put \(\displaystyle{a}=-{16}\) in equation (2), we get
\(\displaystyle\Rightarrow{2}{\left(-{16}\right)}+{b}=-{32}\)
\(\displaystyle\Rightarrow{b}={0}\)
Now put \(\displaystyle{a}=-{16}\), \(\displaystyle{b}={0}\) & \(\displaystyle{c}={256}\) in equation (1), we get
\(\displaystyle{h}{\left({1}\right)}=-{161}^{{{2}}}+{256}\)

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