The temperature of the air is 20 Celsius when a tourist drops a coin into a wishing well. She hears the splash 5.30s after she drops the coin. If the air resistance is ignored, how deep is the well?
a) 344m b)332m c)137m d)120m e)0.60m
time it takes the coin to fall: h=1/2 g t^2
t=sqrt(2h/g)
time it takes sound to come back: h/vs
given: those times add to 5.2 seconds
5.2=sqrt(2H/g)+h/v
a quadratic: let u=sqrt h so u^2=h
u^2/v+u(sqrt2/g)-5.2=0
u^2+u*vsqrt(2/g)-5.2v=0
calculate v for that temerature, then use the quadratic equation to solve for u (and square it for h)
i dont really understand the explanation, so which would be the answer
To find the depth of the well, we can use the formula for free-fall motion:
d = (1/2) * g * t^2
Where:
d is the depth of the well,
g is the acceleration due to gravity (9.8 m/s^2),
and t is the time it takes for the coin to fall.
Given that the time it takes for the tourist to hear the splash is 5.30 seconds, we can substitute this value into the formula as follows:
d = (1/2) * (9.8 m/s^2) * (5.30 s)^2
Calculating this expression will give us the depth of the well. Let's calculate it:
d = (1/2) * (9.8 m/s^2) * (5.30 s)^2
d = 0.5 * 9.8 m/s^2 * 28.09 s^2
d = 4.9 * 28.09 m
d ≈ 137.441 m
Therefore, the depth of the well is approximately 137 meters. So, the correct option would be c) 137m.