Aprojectile travels above level ground. the initial vertical component of its velocity is 0.90 m/s and the initial horizontal component of its velocity is 0.60 m/s.
calculate the time the projectile remains in the air, and calculate the horizontal distance the projectile travels.

at maximum height the final velocity will be 0 
using v=u+at and resolving vertically we get 
v=0.6+(-9.81)t 
v=0.6-9.81t 
0=0.6-9.81t 
9.81t=0.6 
t=0.6/9.81 
t=0.061 to 3sf 

Now we need to resolve horizontally to find the horizontal distance 
using s=ut+1/2at^2 
However we now need the total time taken for the projectile travel and return to the ground. We can assume the time taken for the projectile to reach its maximum height and return to the ground is the same therefore 
the total time is 2 x 0.061=0.122seconds. They'll be now horizontal acceleration in this case scenario therefore 

Hence s=ut+1/2at^2 
since a=0 
s=ut 
s=0.6 x 0.122 
s=0.073m