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?

Question

anonymous · Answer

0 = (t -6)(t + 2)
=> t=6 and t=-2
Correct solution is t=6 because negative time is not possible

anonymous · Answer

Erm.. I got 1.29,-9.29

anonymous · Answer

you can rewrite the equation as 0 =t2 ± 4t ± 12

anonymous · Answer

You're right mkuma.. I just scratched it out :/

anonymous · Answer

t^2 - 4t -12 =0
(t-4)*(t+3) =0