Algebra II Problem, involving calculating distance.

Question

Problem: A stone is dropped into a deep, well. You don't see the stone hit the water, but you hear it splash 4 seconds after it's dropped. What is the distance from the top of the well to the surface of the water?

Key Fact #1: Sound travels at 343 m/s.

Key Fact #2: The height of an object t seconds after it is thrown from an initial height ho with an upward velocity of vo m/s is given by the equation h(t) 4.9t2 vot ho

Answer 1

I grew up in Baltimore. Now I am retired and live in Florida.

I also teach physics in high school now.

 

Time is takes the sound to reach your ears is d/343

Time is takes to fall is 4-d/343=T

d = 4.9T² =4.9(4-d/343)²

 

You have a quadratic equations to solve for d

 

d= 4.9(4-d/343)^2. I will leave this for you to finish

 

I am available for online tutoring

 

Answer 2

Two things have to happen. First, the stone has to travel to the water. Then the sound has to travel back. The round trip takes four seconds. If the distance to the water is H, you can find the time it takes the stone to fall by solving

H = v_ot₁+4.9t₁^2

They don't explicitly tell you v_o, but you should be able to figure it out. (Hint: the stone is dropped, not thrown)

You can find the time it takes for the sound to come back out of the well with

speed of sound = H/t₂ 

You know t₁+t₂=4

The rest should be the usual algebra.