A rock is dropped from a tall building. If the rock starts at rest, and the origin of the coordinate system is taken at the pinnacle of the building with positive direction taken to be upward direction, after 5.18 seconds:

a) What is the displacement of the rock?

b) What is the velocity of the rock?

a) y=-131.47 m : if the coordinate system is taken at the pinnacle of the building with positive direction taken to be upward direction.

b) v=-50.764m/s:The minus sign indicates the direction of the speed that is down

Explanation:

Conceptual analysis

We apply the free fall formula for position (y) and speed (v) at any time (t).

y = v₀*t + ½ g*t² Equation 1

v=v₀+g*t Equation 2

y: The vertical distance the ball moves at time t

v₀: Initial speed in m/s

g = acceleration due to gravity in m/s²

v = Speed the ball moves at time t

Known information

v₀=0

t=5.18 s

g=9.8 m/s²

Development of problem

a) We replace t in the equations (1) to y (5.18s):

y = o+½ * 9.8*5.18² = 131.47m

y=-131.47 m : if the coordinate system is taken at the pinnacle of the building with positive direction taken to be upward direction

b) We replace t in the equations (2) to v (5.18s):

v=o+9.8*5.18=50.764m/s

v=-50.764m/s:The minus sign indicates the direction of the speed that is down