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anonymous
 one year ago
An object is launched at 19.6 meters per second (m/s) from a 58.8meter 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 @zepdrix
anonymous
 one year ago
An object is launched at 19.6 meters per second (m/s) from a 58.8meter 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 @zepdrix

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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\[\large\rm s(t) = –4.9t^2 + 19.6t + 58.8\]Mmm k so this is the same as the third part of the last problem. We care about the time, t, at which the height of the function, s(t), is zero. So we'll let s(t) be zero and then plug all the goodies into the quadratic formula.\[\large\rm 0 = –4.9t^2 + 19.6t + 58.8\]Plug em in! :O\[\rm t=\frac{b\pm\sqrt{b^24ac}}{2a}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441258743147:dw

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0What's under the square root? :O It should be 1500 or something
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