Physics help please!

Question

An elevator, starting at rest on the ground level of a building, climbs vertically upward with a constant acceleration of 0.5 m/s^2, as measured from the ground. A person (not in the elevator) is running directly away from the building at a constant speed of 8 m/s. Determine the speed of this person, as measured by someone in the elevator, at the moment the elevator is 36 meters from the ground.

Answer 1

Speed = 16.4 m/s

You have to find the speed of the elevator when it is 36 feet above the ground. You can use formulas from kinematics or integration. Then you need to draw the two velocity vectors: the elevator's and the runner's. The resultant is relative speed.

Actually, the vector from the origin to the elevator + the vector from the elevator to the runner = the vector of the runner.

So what you really want is the vector length (the speed) of  the vector from the elevator to the runner . That's the same size as the resultant of the two vectors I stated before because the quadrilateral is a rectangle.