Problema Solution
an object is launched at 19.6 meters per second (m/s) from a 58.8-meter platform. The equation for the object's height(s) at time(t) seconds after launch is s(t)= -4.9t^2+19.6t+58.8, where is (s) in meters. When does the object strike the ground?
Answer provided by our tutors
s(t)= -4.9t^2+19.6t+58.8=0
So
-4.9t^2+19.6t+58.8=0
t= 6
So after 6 seconds it strike the ground