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# A spelunker drops a stone from rest into a hole. The speed

### Customer Question

A spelunker drops a stone from rest into a hole. The speed of sound is 343 m/s in air, and the sound of the stone striking the bottom is heard 1.50s after the stone is dropped. How deep is the hole?

[Acceleration of gravity G = 9.8 m/s2]
v = v0 + at
h = ½(v0 + v)t
h = v0t + ½ at2
v2 = v02 + 2ah

[Distance “h” substituted for “x” in this case (h = height)]

Thank you.
Submitted: 7 years ago.
Category: Homework
Expert:  muvee replied 7 years ago.

v0=0 since drop the stone from rest, then

h=½ g(t1)² where t1 is the time required for stone reach the bottom

Hence, t1=sqrt(2h/g)

the time required for sound reach to ear from bottom is h/343

So sqrt(2h/g)+(h/343)=1.5

sqrt(h/4.9)=1.5-(h/343) both sides square and get

h/4.9=2.25-(3h/343)+h^2/343^2

h^2/343^2 - (3/343 + 1/4.9)h+2.25=0 solve this quadratics equation and get

h=10.5764 m

Customer: replied 7 years ago.

Original Question was: A spelunker drops a stone from rest into a hole. The speed of sound is 343 m/s in air, and the sound of the stone striking the bottom is heard 1.50s after the stone is dropped. How deep is the hole?

v0=0 since drop the stone from rest, then
h=½ g(t1)² where t1 is the time required for stone reach the bottom

Hence, t1=sqrt(2h/g)

the time required for sound reach to ear from bottom is h/343

So sqrt(2h/g)+(h/343)=1.5

sqrt(h/4.9)=1.5-(h/343) both sides square and get

h/4.9=2.25-(3h/343)+h^2/343^2

h^2/343^2 - (3/343 + 1/4.9)h+2.25=0 solve this quadratics equation and get

h=10.5764 m

NEW QUESTIONS:

1) How does sqrt(2h/g) become sqrt(h/4.9)? g=9.8m/s.s
2) How is the quadratic equation solved?

Expert:  muvee replied 7 years ago.