Problem

At \(t\) seconds after launch, an object's height is \(h(t)~=~-16t^{2}~+~64t\). Use the displayed graph to set the height equal to \(0\), solve for \(\text{both}~\text{roots}\), and identify the requested time after launch.

A scaled graph of h of t equals negative sixteen t squared plus sixty-four t from zero through four seconds. The parabola begins at height zero, reaches height sixty-four at two seconds, and returns to the ground-height target. The intersection times are not annotated. — The motion model and ground-height target are shown on scaled axes without labeled intersections.

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What this problem is really about

Reaching the ground means the height model has output zero, so the needed times are the roots of a quadratic equation. We’ll factor the model, solve for both graph intersections with the ground, and then use the phrase “after launch” to distinguish the return time from the initial launch moment.

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Equation: −16t²+64t=0; roots: t=0 and t=4; requested time: t=0 seconds.

Equation: −16t²+64t=0; roots: t=0 and t=4; requested time: t=4 seconds.

Equation: −16t²+64t=64; roots: t=2 and t=2; requested time: t=2 seconds.

Equation: −16t²+64t=0; roots: t=0 and t=2; requested time: t=2 seconds.

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At \(t\) seconds after launch, an object's height is \(h(t)~=~-16t^{2}~+~64t\). Use the displayed graph to set the height equal to \(0\), solve for \(\text{both}~\text{roots}\), and identify the requested time after launch.

What this problem is really about

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