I need help solving this indefinite integrals problem? I am doing indefinite integrals homework, and this problem popped up. I hate to post on here without any personal insight on the problem, but I really have no idea on how to approach this.I do not know what to do with the information given. Any insight on how to solve the problem is what I am looking for, not specifically an answer. Thanks for all the help in advance. 
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Kaitlyn drops a stone into a well. Approximately $4.61$sec later, she hears the splash made by the impact of the stone in the water. How deep is the well? (The speed of sound is approximately $1128$ ft./sec. Round your answer to the nearest foot.)
 A: You can use physics to solve this actually:
The speed of sound is $$1128\ \text{ft/sec}$$
And the total time it took to travel the height of the well is $4.61$ seconds.
Now:
$$x(t) = -16(t_1)^2 + h_0 = 0$$
$$h_0 = 1128*t_2$$
And $$t_1 + t_2 = 4.61$$
So,
$$t_1  = \sqrt{\frac{h_0}{16}}$$
$$t_2 = \frac{h_0}{1128}$$
So, we have that:
$$\sqrt{\frac{h_0}{16}} + \frac{h_0}{1128} = 4.61$$
Solving,
we find that $h_0 = 301.00 = 301\ \text{feet}$
hope this helped.
A: Here's a hint:
The total time between dropping the stone and hearing the splash can be broken down into two parts:


*

*The time it takes for the stone to hit the water after being released from your hand ($t_1$)

*The time it takes for the sound of the splash to reach your ear ($t_2$)


You can write down an equation in terms of the height of the well, $h$, for each part, in terms of the time consumed on each part.  You can write an equation for the known total time, $T$, in terms of the individual times.
Three equations, three unknowns ($t_1, t_2, h$).
A: So, we know that $9.8\cdot \frac{1}{2}\cdot (\Delta(t_{1}))^2 = v_{sound}\cdot \Delta(t_{2})$ and $\Delta(t_{1}) + \Delta(t_{2}) = 4.61$.
So, $$\frac{4.9}{v_{sound}}\cdot\Delta(t_{1})^2+\Delta(t_{1}) - 4.61 = 0.$$
So, $\Delta(t_{1}) = -\frac{v_{sound}}{9.8} + \frac{\sqrt{1+19.6\cdot 4.61/v_{sound}}}{9.8/v_{sound}} = 4.34 s$, where $v_{sound} = 344 m/s$. 
So, $h = \Delta(t_{2})\cdot (344 m/s) = (4.61-4.34 s)\cdot (344 m/s) = 92.4 m$. Now, this is equivalent to $303$ feet (if you dare not use SI).
