Problema Solution
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 provided by our tutors
the object strikes the ground when
s(t) = -4.9t^2 + 19.6t + 58.8, for some t (t>0)
by solving the equation
-4.9t^2 + 19.6t + 58.8 = 0
we find
t = 6 s
click here to see the step by step solution of the equation
the object strikes the ground after 6 seconds.