A ball is launched vertically upward from ground level with an initial velocity of 30 m/s. How much time does it take before it lands on the ground?

This problem has to be broken down into two parts: the trip up and the trip down. Using the equation Vf^2=Vi^2+2(a)(d) we can substitute the following values:
Vf= 0m/s
Vi= 30m/s
a= -10m/s (use 9.81 for regents physics)
d=?
t=?

we can now solve the equation-
(0m/s)=(30m/s)+(-10m/s^2)(t)

giving us t=3sec

this is the time for the trip up only, however we know that the trip up= the trip down, so the time is 6 seconds.

Hope this helps!