 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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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
$$\displaystyle{h}{\left({1}\right)}=-{161}^{{{2}}}+{256}$$