Question

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? ...?

Answer 1

from the information above, we can find out that

-4.9 t^2 + 19.6 t + 58.8 = 0 (divide all with -4.9)

t^2 + 4t + 12 = 0

(t-6) (t+2) = 0

t = 6s

hope this helps