Question

(please show work)

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:

$s=ut+\frac{1}{2}gt^2$

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:

$s=0+\frac{1}{2}(-9.8)(10)^2=-490 m$

where the negative sign means the displacement is downward.