Calculate the height from the speed of sound

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SUMMARY

The discussion focuses on calculating the depth of a well based on the time it takes for a stone to hit the water and the sound of the splash to be heard, totaling 1.63 seconds. The speed of sound in air is given as 343 m/s. The kinematic equation used is Δh = V0t + 1/2gt², with the initial velocity (V0) set to zero. The solution involves setting the distance fallen equal to the distance sound travels, leading to a quadratic equation that can be solved using the quadratic formula.

PREREQUISITES
  • Understanding of kinematic equations, specifically Δh = V0t + 1/2gt²
  • Knowledge of the speed of sound in air (343 m/s)
  • Familiarity with solving quadratic equations
  • Basic principles of free fall and sound propagation
NEXT STEPS
  • Study the derivation and application of kinematic equations in physics
  • Learn how to apply the quadratic formula to solve equations
  • Explore the relationship between distance, time, and speed in sound propagation
  • Investigate the effects of air temperature on the speed of sound
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and sound, as well as educators looking for practical examples of these concepts in action.

Julio Cesar
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Homework Statement


You drop a stone down a well. You hear the 'splash' 1.63s later. How deep is the well? The speed of sound in air is 343m/s.


Homework Equations


Kinematic equation: Δh=V0t+1/2gt2

Also, I've narrowed the problem down to the total time of 1.63s is equal to...
1) the time the stone leaves the hand and hits the water bellow. This is t1
2) The time it takes for the sound to rise from the water and be heard above. This is t2

So 1.63s = t1 + t2

The Attempt at a Solution



Now since the initial velocity is zero. We can just say that Δh=1/2gt12

And the speed of sound is v=meters/sec (since I want the distance I can call meters displacement.) Right?

So now I have two equations that I can rearange for time (t1 and t2

And I also have the full total time of Ttotal which is equal to the sum of t1 + t2

But my question is this...I need to find the displacement (or height) but I have three unknowns (both time values and the height) but I also have two equations. So my brilliant self did this...

1.63s = √2(h)/g + h/vs

I tried to solve for "h" and that ended horribly. So now I'm stuck. I know I can get this. I desperately want to solve this now to prove to myself that I am not stupid. So I've been stuck for over a few hours (taking breaks of course) and going over it in my head as to how to either find one of the time values or the height altogether.

Please if someone can just give me a little tip. I don't want the answer nor a plug and chug final equation. I just need a little math tip as to what to do next (or what am I doing wrong)...

Thanks for any and all help, ya'll.
 
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You could solve that equation by taking the radical to one side, other terms to the other and squaring both sides. But the easier approach is to start with
distance down = distance up
1/2gt² = vt₂
1/2gt² = v(1.63 - t)
which has only one unknown.
 
You have already done the hard part. Nice job! Now...

Let y = √h, y2 = h, and solve your equation using the quadratic formula.
 
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