Math 1050-1
Homework Set 5, Problem 5 Solution
−gt2
Problem 5 gives us the equation d =
. We notice that no matter what t value we use, this gives
2
us a negative d value. The negative simply means a direction. The person drops the stone, so if we think
of them stading at the origin, the stone goes down |d| units. Since we only want the distance and we don’t
care about the direction, we will use the equation
gt2
2
Statement of the problem: You drop a rock into a deep well. You can’t see the rock’s impact at the
bottom, but you hear it after t seconds. What is the depth of the well?
d = |d| =
We know the time t has two components to it: the time it takes the stone to drop and the time it take
for the sound to come back to us. Let t0 be the time it takes for the stone to drop, and t1 be the time it
takes for the sound to reach us. Then we have the equation:
t = t0 + t1
(1)
We also have the depth equation given to us initially. Substituting our value t = t0 in this equation we
have:
gt20
(2)
2
We now need an equation for the sound. We know the speed of the sound is c, and the distance it traveled
to get to us is d and the time it took was t1 . So we have:
d=
d = ct1
(3)
We can now solve for t0 and t1 in the equations (2) and (3), and substitute them into equation (1) and
solve for d.
We have:
s
2d
(4)
t0 =
g
t1
=
t
=
d
c
s
(5)
2d d
+
g
c
(6)
There are two ways of solving for d at this point.
Method 1
Isolate the square root on one side of the equation, and then square both sides of the equation to get rid
of the square root. Next combine like terms and write the equation as a quadratic in d.
t
d
t−
c
2
d
t−
c
2t
1
t2 − d + 2 d2
c
c
1 2
−2t 1
d + t2
d +
−
c2
c
c
=
s
2d d
+
g
c
(7)
=
s
2d
g
(8)
=
=
2d
g
2
d
g
= 0
(9)
(10)
(11)
Now we can use the quadratic equation to solve for d. We will have
a
=
b =
c
=
So we have:
d=
1
−( −2t
c − c) ±
1
c2
−2t 1
−
c
c
2
t
q
1 2
1 2
( −2t
c − c ) − 4( c2 )t
2
c2
Now the question is, do we want the positive square root or the negative. We know if t = 0 then d = 0,
so what happens when we have t = 0?
q
−(− 1c ) ± (− 1c )2 − 0
d=
2
c2
So we see we want the negative square root. Our answer is then:
q
1
1 2
1 2
−
)
−
( −2t
−( −2t
c
c
c − c ) − 4( c2 )t
d=
2
c2
For problem 5, simply substitute your value for t, and then let g = 32 feet per second squared, and
c = 1100 feet per second.
Method 2
For the second method, we will see we have a quadratic in
√
d. We have the equation
t
=
s
r 1 √ 2
2 √
d +
d−t
c
g
=
0
We see this is a quadratic equation in
2d d
+
g
c
√
d. We have
a
b
c
So we have
−
√
d=
q
2
g
=
1
c
r
2
g
= −t
=
rq
± ( g2 )2 − 4( 1c )(−t)
2
c
Once again we need to determine which square root we should have. So we do the same trick. When
t = 0 we should have d = 0. When t = 0 we have
rq
q
2
( g2 )2 − 0
−
±
g
√
d=
2
c
So we see we need the positive square root. Now square both sides of the equation to get our value for
d. Our answer is:
rq
2
 q
2
2 2
1
(
±
)
−
4(
)(−t)
−
g
g
c



d=
2


c
Once again, for problem 5, simply substitute your value for t, and then let g = 32 feet per second squared,
and c = 1100 feet per second.