please help...A stone is droped down a well and 5 second later the sound of the splash is heard. if the velocity of sound is 1120 feet per second, what is the depth of the well? ..answer 353 feet thanks....

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please help...A stone is droped down a well and 5 second later the sound of the splash is heard. if the velocity of sound is 1120 feet per second, what is the depth of the well? ..answer 353 feet thanks....

Physics
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did you use the equation \[v = x \div t\] where x is the depth , t is the time and v is the velocity of sound and then the depth \[x= v \times t\]
if x=vt=1120x5=5600 feet, the answer should be 353 feet
Let: g = 32.174ft/s T_1 = time for the stone to reach well bottom T_2 = time for sound to reach top of well The height going down will equal the height going up...but the times for each will be different since they are moving at different rates. Therefore, you can simply equate the heights: \[h=\frac{1}{2}{(32.174ft/s^2)}T_1=(1120ft/s)(T_2)\]and you know the sum of the times equals 5 seconds:\[T_1+T_2=5s\] Now you have two equations with two unknowns. To solve, rearrange the second equation and substitute it back into the first equation: \[T_2=5s-T1\]\[h=\frac{1}{2}{(32.174ft/s^2)}T_1=(1120ft/s)(5s-T1)\] Solving for T1 you get 4.68477s. Now you know how long it took the stone to reach the bottom of the well. Therefore, calculate the height using the first kinematic equation above again plugging in T1: \[h=\frac{1}{2}{32.174ft/s^2}(4.68477s)=353ft\]

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yehey great job,,thank u very much
you're welcome :)
I just noticed that I mistakenly left off the fact that the T1s should all be squared in the height equations. The calculations are correct but I just didn't type them out right...and I'm not typing it all again :) Anyway, those parts of the equations are based on the kinematic equation: \[d=V0t+12gt^2\]where V0 is obviously 0 since the stone starts its fall at 0 m/s.
yes i noticed that and thats ok....lol thnk u very much again
No problem :)

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