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Neglecting air resistance, the distance (d) that an object fallsvaries directly as the square of the time (t) it has been falling.If an object falls 64 feet in 2 seconds, determine the distanceit will fall in 6 seconds.

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d(t)=at^2

where a is a constant, we can find a with the first statement:

"an object falls 64 feet in 2 seconds"

so:


\begin{gathered} 64=a(2)^2 \\ a=(64)/(2^2) \\ a=16 \end{gathered}

the complete equation is


d(t)=16t^2

now "determine the distance for 6seconds"

so, replace t=6 and find d


\begin{gathered} d(t)=16(6)^2 \\ d(t)=576ft \end{gathered}

the distance is 576ft

User OGrandeDiEnne
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