I did some more thinking about the problem and think I’ve figured it out. Here’s the solution. After that I’ll explain the components in detail and provide commentary about my thoughts.
In[]:=
equations=r''[t]==
standard acceleration of gravity
PHYSICAL CONSTANT
["Value"],rt-
0.2718`
s
-r[t]==
2.718
m
;
In[]:=
DSolve[equations,r[t],t]
DSolve
:To avoid possible ambiguity, the arguments of the dependent variable in 
′′
r
[t]
196133
20000
m/
2
s
,-r[t]+rt+
-0.2718
s

2.718
m
 should literally match the independent variables.
Out[]=
DSolve
′′
r
[t]
196133
20000
m/
2
s
,-r[t]+rt+
-0.2718
s

2.718
m
,r[t],t
I don’t know how to fix this.
In[]:=
DSolver''[t]==
standard acceleration of gravity
PHYSICAL CONSTANT
["Value"],r[t],t
Out[]=
r[t]

1
+t

2
+
2
t
196133
40000
m/
2
s

Maybe using no Quantities will fix the problem.
In[]:=
DSolve[{r''[t]==-9.8,r[t-0.2718]-r[t]==2.718},r[t],t]
DSolve
:To avoid possible ambiguity, the arguments of the dependent variable in {
′′
r
[t]-9.8,r[-0.2718+t]-r[t]2.718} should literally match the independent variables.
Out[]=
DSolve[{
′′
r
[t]-9.8,r[-0.2718+t]-r[t]2.718},r[t],t]
I think this is a delay differential equation. I changed an example from the documentation and get the error.
In[]:=
sol=DSolve[{x'[t]==x[t-1]^2,x[t]==at},x[t],{t,0,2}]
DSolve
:To avoid possible ambiguity, the arguments of the dependent variable in 
′
x
[t]
2
x[-1+t]
,x[t]at should literally match the independent variables.
Out[]=
DSolve[{
′
x
[t]
2
x[-1+t]
,x[t]at},x[t],{t,0,2}]
The example from DSolve, Scope, Delay Differential Equations works when /;t<=0 is added.
In[]:=
sol=DSolve[{x'[t]==x[t-1]^2,x[t/;t<=0]==at},x[t],{t,0,2}]
Out[]=
x[t]
at
t≤0
1
3
(3
2
a
t-3
2
a
2
t
+
2
a
3
t
)
0<t≤1
1
126
(42
2
a
-163
4
a
+686
4
a
t-1176
4
a
2
t
+1064
4
a
3
t
-553
4
a
4
t
+168
4
a
5
t
-28
4
a
6
t
+2
4
a
7
t
)
1<t≤2
Indeterminate
True

The error still exists with condition if t is changed to {t,0,2}.
In[]:=
sol=DSolve[{x'[t]==x[t-1]^2,x[t/;t<=0]==at},x[t],t]
DSolve
:To avoid possible ambiguity, the arguments of the dependent variable in 
′
x
[t]
2
x[-1+t]
 should literally match the independent variables.
Out[]=
DSolve[{
′
x
[t]
2
x[-1+t]
,x[t/;t≤0]at},x[t],t]
Maybe it will work with /;
In[]:=
DSolve
′′
r
[t/;t>=0]
196133
20000
m/
2
s
,-r[t]+rt+
-0.2718`
s

2.718`
m
,r[t],t
DSolve
:To avoid possible ambiguity, the arguments of the dependent variable in -r[t]+rt+
-0.2718
s

2.718
m
 should literally match the independent variables.
Out[]=
DSolve
′′
r
[t/;t≥0]
196133
20000
m/
2
s
,-r[t]+rt+
-0.2718
s

2.718
m
,r[t],t
I don’t get the error when I specify a tmin and a tmax. I don’t know what the max time is so I’ll go for 100 s, which seems reasonable for the time it takes to fall.
In[]:=
DSolve
′′
r
[t/;t>=0]
196133
20000
m/
2
s
,-r[t]+rt+
-0.2718`
s

2.718`
m
,r[t],{t,0,100}
Out[]=
DSolve
′′
r
[t/;t≥0]
196133
20000
m/
2
s
,-r[t]+rt+
-0.2718
s

2.718
m
,r[t],{t,0,100}
Maybe the quantities are making it not evaluate.
In[]:=
DSolve[{
′′
r
[t/;t>=0]-9.8,-r[t]+r[t+-0.2718]2.718},r[t],{t,0,100}]
Out[]=
DSolve[{
′′
r
[t/;t≥0]-9.8,r[-0.2718+t]-r[t]2.718},r[t],{t,0,100}]
In[]:=
DSolve[{
′′
r
[t/;t>=0]-9.8,-r[t]+r[t+-0.2718]2.718},r[t],{t,0,100}]
Out[]=
DSolve[{
′′
r
[t/;t≥0]-9.8,r[-0.2718+t]-r[t]2.718},r[t],{t,0,100}]
I can find how high a stone would fall from it if it took 100 s to fall.
In[]:=
UnitConvert12
g

2

100
s

,"Meters"
Out[]=
196133
4
m
In[]:=
NUnitConvert12
g

2

100
s

,"Kilometers"
Out[]=
49.0333
km
I don’t think the stone fell from a height of
49.0333
km.
Find entities for physical constants in the acceleration class.
In[]:=
EntityValue[EntityClass["PhysicalConstant","Acceleration"],"Entities"]
Out[]=

acceleration associated with cosmological expansion rate
,
MOND constant
,
standard acceleration of gravity

In[]:=
DeleteMissingEntityValue
standard acceleration of gravity
PHYSICAL CONSTANT
,"PropertyAssociation"
In[]:=
C[1]
Out[]=

1
I have the equation
r''(t)=-9.8m/
2
s
, or a more precise value for the standard acceleration of gravity such as
-9.81m/
2
s
. r(t) represents the vertical position, or height at time t. r(0) is the height of the building. Integrating the acceleration r’’(t)=-g gives the velocity
r'(t)=-gt+

1
. Integrating the velocity to determine the position gives
r(t)=-g
2
t
2
+

1
t+

2
, which has a similar form the textbook’s equation of
x=
x
0
+
v
0
t+
1
2
a
2
t
.
Let
t
1
and
t
2
be the times at which the stone passes the top and bottom parts of the window, respectively. For example
t
1
could be 1 second. This means the stone passes the top part of the window 1 second after the stone is dropped from the roof. Then 0.2718 s later at
t
2
the stone passes the bottom part of the window. We have the equation
t
2
=
t
1
+0.2718s
.
I need some help figuring out how to enter the ODE.