Answer:
(A)
[tex]\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s \right )D+t_t^2v_s^2=0[/tex]
(B) D=54.71 m
Explanation:
Free Fall
When a particle is dropped in free air, it starts falling to the ground with an acceleration equal to the gravity. If one wanted to know the height of launching, it can indirectly be measured by the time it takes to reach the ground by the formula
[tex]\displaystyle D=\frac{gt^2}{2}[/tex]
Solving for t
[tex]\displaystyle t=\sqrt{\frac{2D}{g}}[/tex]
If we are taking into consideration the time we can hear the sound it makes when hitting the ground (or water in this case), we must also consider the speed of the sound for the time it takes to reach back our ears. That time can be computed from the basic equation for the speed
[tex]\displaystyle t=\frac{D}{v_s}[/tex]
(A)
The total measured time is the sum of both times and it's given as [tex]t_t=3.5\ seconds[/tex]
[tex]\displaystyle t_t=\sqrt{\frac{2D}{g}}+\frac{D}{v_s}[/tex]
From this equation we'll manage to compute D
First, we isolate the square root
[tex]\displaystyle \sqrt{\frac{2D}{g}}=t_t-\frac{D}{v_s}[/tex]
Let's square both sides
[tex]\displaystyle \frac{2D}{g}=t_t^2-2t_t\frac{D}{v_s}+\frac{D^2}{v_s^2}[/tex]
Multiplying by [tex]v_s^2[/tex]
[tex]\displaystyle \frac{2Dv_s^2}{g}=t_t^2v_s^2-2t_tDv_s+D^2[/tex]
Rearranging and factoring
[tex]\boxed{\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s\right )D+t_t^2v_s^2=0}[/tex]
Now, let's put in numbers:
[tex]g=9.8\ m/s^2,\ v_s=345\ m/s,t_t=3.5\ sec[/tex]
[tex]\displaystyle D^2-\left (\frac{2(345)^2}{9.8}+2(3.5)(345)\right )D+(12.25)345^2=0[/tex]
Computing all the coefficients:
[tex]\displaystyle D^2-26,705.82D+1,458,056.25=0[/tex]
Solving for D, we have two possible solutions:
[tex]D=54.71,\ D=26,651.11[/tex]
The second solution is called "extraneous", since it comes from squaring an equation, which can introduce non-valid (or external) solutions. It's impossible, given the conditions of the problem, that the well could be 26.5 km deep. So we'll keep the only solution as.
D=54.71 m
Let's prove our calculations by computing both times:
[tex]\displaystyle t_1=\sqrt{\frac{2(54.71)}{9.8}}=3.34\ sec[/tex]
[tex]\displaystyle t_2=\frac{54.71}{345}=0.16\ sec[/tex]
We can see their sum is 3.5 seconds, 3.34 of which were taken to reach the bottom of the well, and 0.16 sec took the sound to reach the top.
