I got this question from a website:
An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t2 + 19.6t + 58.8, where s is in meters. When does the object strike the ground?
What is the height (above ground level) when the object smacks into the ground? Well, zero, obviously. So I'm looking for the time when the height is s = 0. I'll set s equal to zero, and solve:
0 = –4.9t2 + 19.6t + 58.8
0 = t2 – 4t – 12
0 = (t – 6)(t + 2)
Then t = 6 or t = –2. The second solution is from two seconds before launch, which doesn't make sense in this context. (It makes sense on the graph, because the line crosses the x-axis at –2, but negative time won't work in this word problem.) So "t = –2" is an extraneous solution, and I'll ignore it.
So what would happen if both answers were positive. What would be the answer?
2007-03-27
00:41:09
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6 answers
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asked by
Anthony
4
in
Mathematics