(a) the brick is in free fall, so it is moving by uniformly accelerated motion, with constant acceleration [tex]g=9.81 m/s^2[/tex]. The velocity of the brick at time t is given by
[tex]v(t) = v_0 +gt[/tex]
where v0 is the initial velocity of the brick (zero, in this case). Using this formula, we can calculate the velocity at t=4.0 s:
[tex]v(4.0s)=gt=(9.81 m/s^2)(4.0 s)=39.2 m/s[/tex]
(b) The distance covered by the brick in accelerated motion after a time t is given by
[tex]S= \frac{1}{2} gt^2[/tex]
If we substitue t=4.0 s, we find how far the brick falls during this time:
[tex]S= \frac{1}{2} (9.81 m/s^2)(4.0s)^2=78.5 m[/tex]
