c. -98 m/s
The motion of the rock is a uniformly accelerated motion (free fall) with constant acceleration [tex]g=-9.8 m/s^2[/tex] (negative since it is downward). Therefore, we can find its velocity using the following suvat equation
[tex]v=u+gt[/tex]
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
[tex]v=0+(-9.8)(10)=-98 m/s[/tex]
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:
[tex]s=ut+\frac{1}{2}gt^2[/tex]
where
s is the displacement
u is the initial velocity
g is the acceleration of gravity
t is the time
Substituting:
u = 0
[tex]g=-9.8 m/s^2[/tex]
t = 10 s
We find the rock's displacement:
[tex]s=0+\frac{1}{2}(-9.8)(10)^2=-490 m[/tex]
where the negative sign means the displacement is downward.
