A coin is dropped in a 15.0 m deep well. A)How long does it take the coin to hit the surface of the water?

B)How much time would it take to hear the splash after dropping the coin if the speed of sound is 340. m/s?

t = 1.75

t = 0.04

Explanation:

a)

For part 1 we want to use a kenamatic equation with constant acceleration:

X = 1/2*a*t^2

isolate time

t = sqrt(2X / a)

Plugin known variables. Acceleration is the force of gravity which is 9.8 m/s^2

t = sqrt(2*15m / 9.8m/s^2)

t = 1.75 s

b)

The speed of sound travels at a constant speed therefore we don't need acceleration and can use the equation:

v = d / t

isolate time

t = d / v

plug in known variables

t = 15m / 340m/s

t = 0.04 s

0.19 m/s and 0.45 m/s

small = small current large mag large current