An archer stands 40.0m from the target. if the arrow is shot horizontally with a velocity of 90.0 m/s, how far above the bull's eye must she aim to compensate for gravity pulling her arrow downward?

Ok, first we need to have in mind that to findthe distance we need to find also the time then we use the kinematic equations in the x-direction 

delta x = Vox*t + 1/2*a*t^2 
since there is no horizontal acceleration, a = 0 m/s/s 
delta x = Vox*t 
40.0 m = 90 m/s * t 
t = 4/9 s 

Now using this time you can calculate the vertical displacement due to gravity. 
delta y = Voy*t + 1/2*g*t^2 
since there is no initial vertical velocity, Voy = 0m/s 
delta y = 1/2*g*t^2 
delta y = 1/2*(-9.81 m/s/s)*(4/9 s)^2 
solve for delta y and you will have the answer

Is a  vector quantity  that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.