Indigo28

An object is thrown from an initial height of 0 feet with an initial velocity of 160 feet per second. Solve for when it will return to the ground. Show your work and explain the steps you used to solve.

romsek

Since its starting and ending altitudes are the same, 0 feet, the time it returns to the ground is twice that of when it reaches the top of its trajectory.

At the top of its trajectory, its vertical speed is 0 ft/s.

That should get you started.

Country Boy

It would be better for you to "show your work and explain the steps you used"! Are you given the formula "s= (a/2)t^2+ (v0)t+ s0" (v0 is the initial speed, s0 is the initial height)? The only other way I can think of would
use Calculus.

In this problem v0= 160 and s0= 0 while a is the downward acceleration is 32 feet per second per second. S+ o it will return to the ground when (-32/2)t^2+ 160t= 0. That equation should be easy to solve. It has two solutions but it should be clear which is the one you want!