Use the position function s(t)=-16t^{2}+v_{0}t+s_{0} for free-fa

Step 1
a) The position function:
in order to determine this equation we have to determine ${\displaystyle v_0\text{and}{s}_0}$:
the coin is dropped with no initial velocity (free falling object) so we conclude that ${\displaystyle v_0=0}$.
When ${\displaystyle t=0}$ have ${\displaystyle s\left(0\right)=s_0=1362}$ (because the coin was dropped from the top of the building with a height ${\displaystyle =1362}$)
${\displaystyle s\left(t\right)=-16t^2+v_{0}t+s_0}$
${\displaystyle v_0}$ is the initial velocity (when ${\displaystyle t=0}$)
${\displaystyle s_0}$ is the initial position (when ${\displaystyle t=0}$)
(these concepts are from physics!)
we conclude that: the position function ${\displaystyle =s\left(t\right)=-16t^2+1362}$
Step 2
a) the velocity function ${\displaystyle =\frac{d\left(s\left(t\right)\right)}{dt}}$
$v(t)=\frac{d(s(t))}{dt}=-32t$
Step 3
b) the average velocity ${\displaystyle =\frac{s\left(2\right)-s\left(1\right)}{2-1}=s\left(2\right)-s\left(1\right)}$
the average velocity ${\displaystyle =s\left(2\right)-s\left(1\right)=-16\times 4+1362-(-16\times 1+1362)=-48}$
Step 4
c) the instantaneous velocity at 2 is v(2)
the instantaneous velocity at 1 is v(1)
${\displaystyle v\left(2\right)=-32\times 2=-64}$
${\displaystyle v\left(1\right)=-32}$
Step 5
d) time required to the coin to reach the ground:
when coin reach the ground we have ${\displaystyle s\left(t\right)=0}$
so we must solve this equation to determine the time
${\displaystyle s\left(t\right)=0}$
${\displaystyle \Rightarrow -16t^2+1362=0}$
${\displaystyle \Rightarrow -16t^2=-1362}$
${\displaystyle \Rightarrow t^2=\frac{1362}{16}=85.125}$
${\displaystyle \Rightarrow t=\sqrt{85.125}\approx 9.23}$
Step 6
e) plug the time when the coin reaches the ground into the velocity equation: $v(t)=-32t$
${\displaystyle v\left(\sqrt{85.125}\right)\approx v\left(9.23\right)=-32\times 9.23=-295.36}$
Answer:
a) $s(t)=-16t^2+1362,v(t)=-32t$
b) the average velocity ${\displaystyle =-48}$
c) ${\displaystyle v\left(1\right)=-32,v\left(2\right)=-64}$
d) ${\displaystyle t\approx 9.23}$
e) ${\displaystyle t\approx -295.36}$

Good grief, feet and inches and yards. That is an old textbook with
${\displaystyle g=\frac{-32ft}{secon{d}^2}so\frac{g}{2}=16}$
anyway ${\displaystyle s=\left(\frac{1}{2}\right){ft}^2+Vot+Ho}$
so a) ${\displaystyle s=-16t^2+0+1362}$
${\displaystyle \frac{ds}{dt}=velocity=-32t+0}$
and of course acceleration ${\displaystyle =\frac{d^{2}s}{{dt}^2}=-32}$
b) find out how far it moved in that one second and divide by one second
${\displaystyle s\left(2\right)=-16\cdot 4+1362}$
${\displaystyle s\left(1\right)=-16\cdot 1+1362}$
difference ${\displaystyle =-48}$ ft in one second so average speed ${\displaystyle =-48}$ ft/second beteen ${\displaystyle t=1\text{and}t=2}$
c) ${\displaystyle v\left(1\right)=-32\left(1\right)=-32f\frac{t}{s}}$
${\displaystyle v\left(2\right)=-32\left(2\right)=-64f\frac{t}{s}}$
d) when will ${\displaystyle s=0}$?
${\displaystyle 0=-16t^2+0+1362}$
${\displaystyle 16t^2=1362}$
${\displaystyle t^2=681}$
${\displaystyle t=26.1}$ seconds
e) ${\displaystyle -16\left(26.1\right)=-32\left(26.1\right)}$