Question

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?

Answer 1

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

Step-by-step 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