Glenda drops a coin from ear level down a wishing well. The coin falls a distance of 14.80 m before it strikes the water. If the speed of sound is 343 m/s, how long after Glenda releases the coin will she hear a splash?

How long does it take for the coin to fall to the water? distance = 1/2gt^2. Solve for t.

How long does it take for the sound to travel back to Glenda's ear? distance = vt. solve for t. Add the time it takes for the coin to fall to the water and the time for the sound to get to the ear. The total time is ??

3.06 seconds

I think the last Anonymous poster (12 October) neglected to take the square root of t1 (the time it takes the coin to hit the water), the formula for which would be written initially as [(14.8•2)/9.8]=t^2; taking the square root of the left side (3.02) would yield t=1.74 seconds. This, added to the 0.043 seconds it would take for the sound of the splash to travel back to Glenda, equates 1.783 seconds, which would be the answer.

To find the answer, we need to calculate the time it takes for the sound of the splash to reach Glenda after she releases the coin.

We can use the equation: distance = speed × time.

In this case, the distance the sound travels is 14.80 m. Since the sound travels at a speed of 343 m/s, we can plug in these values to find the time:

14.80 m = 343 m/s × time

Dividing both sides of the equation by 343 m/s gives us:

time = 14.80 m / 343 m/s

Calculating this division, we find that the time it takes for the sound of the splash to reach Glenda is approximately 0.043 seconds.