What goes up, must come down. It may not have been written but you were also given the acceleration due to gravity. (9.8m/s^2). Gravity is first acting against the object being thrown up, then it will reach a point where v = 0. Then the object will return down as gravity works with the object.
How do you properly solve for the time till an object hits the ground when its thrown upwards off a cliff? With only The initial velocity and the hieght of the cliff.
Find vertical info.
d = vt + 1/2(atsquared)
you have v, d is the height of the cliff (a negative value) and you have a(-9.8m/ssquared which is gravity), solve for t using quadratic formula. The time vertically is the same as time horizontally but since you are only looking for time the horizontal info is not necessary.
No because you are looking for the time when the ball reaches the ground. Therefore you have to ask yourself what will be the vertical displacement for the time you are looking for. It is -d. This makes sense because althought the ball will go up for a certain d, it will also come back down for that same d just in the opposite direction and will continue moving until it has fallen the height of the cliff (-d)