Question 111135
{{{s=-16t^2+v[0]t+s[0]}}}
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{{{s=-16t^2+64t+25}}}: substituting the values of {{{v[0]=64 fps}}} and {{{s[0]=25feet}}}
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After 1 second, {{{s=-16(1)^2+64(1)+25=-16+64+25=73}}}
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{{{s=-16t^2+64t+25}}} would graph as a parabola opening downward, and the maximum height would occur at the vertex of the parabola.  The t-coordinate of the vertex is given by {{{-b/2a=-64/(2(-16))=2}}}.
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{{{s=-16(2)^2+64(2)+25=-64+128+25=89feet}}}
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{{{graph(400,400,-0.5,6,-10,100,-16x^2+64x+25,25)}}}
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When interpreting the graph, remember that the x-axis is time, and the y-axis is distance.  Notice that the graph crosses the y-axis at 25 feet because that is the value of {{{s[0]}}}.  Also, anything on the graph to the left of the y-axis is meaningless because that would represent negative time values, and anything below the x-axis represents something beneath the ground.
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