Question

(please show work)

A rock is dropped off of a cliff. It falls to the ground in 10 seconds. Answer the following questions assuming that there is no air resistance.


c. What is the instantaneous velocity of the rock just before it hit the ground?


d.Find the rock's displacement for the 10 seconds.​

Answer 1

c. -98 m/s

The motion of the rock is a uniformly accelerated motion (free fall) with constant acceleration
`g=-9.8 m/s^2` (negative since it is downward). Therefore, we can find its velocity using the following suvat equation

`v=u+gt`

where

v is the final velocity

u is the initial velocity

g is the acceleration of gravity

t is the time

For the rock in the problem,

u = 0

So, its velocity at t = 10 s is

`v=0+(-9.8)(10)=-98 m/s`

where the negative sign indicates that the velocity points downward.

d. -490 m

Since the motion is at constant acceleration, we can use another suvat equation:

`s=ut+(1)/(2)gt^2`

where

s is the displacement

u is the initial velocity

g is the acceleration of gravity

t is the time

Substituting:

u = 0

`g=-9.8 m/s^2`

t = 10 s

We find the rock's displacement:

`s=0+(1)/(2)(-9.8)(10)^2=-490 m`

where the negative sign means the displacement is downward.