On hiking up a mountain that has several overhanging cliffs, a climber drops a s

Question

On hiking up a mountain that has several overhanging cliffs, a climber drops a stone at the first cliff to determine its height by measuring the time it takes to hear the stone hit the ground.

A) At a second cliff that is twice the height of the first, the measured time of the sound from the dropped stone is

less than double that of the first  (WHY?)

B) If the measured time is 4.30s

for the stone dropping from the first cliff, and the air temperature is 20?C, how high is the cliff?

d=_____m

C) If the height of a third cliff is three times that of the first one, what would be the measured time for a stone dropped from that cliff to reach the ground?

t=_____s

Explanation / Answer

A)Less than double.

You should add rock travel time to sound travel time. When the distance doubles, the travel time of the rock will not double, it will be less than double. But the sound travel will exactly double with the distance. Here is rough calculations:
100 meters X 2 = 200: 200 / 9.8 = 20.4: sq.root of 20.4 = 4.5 sec (rock travel) + 0.3 sec (sound travel) = 4.8 sec.
200 meters X 2 = 400: 400 / 9.8 = 40.8: sq.root or 40.8 = 6.4 sec (rock travel) + 0.6 sec (sound travel) = 7 seconds.

B)S = ut + 1/2*g*t^2 where S is the height, u is the initial velocity of the stone that is 0 in this case, t is the time taken to reach the ground and g is the acceleration due to gravity which is a constant 10. So you get

S = 0*4.3 + 1/2*10*4.3^2
S = 92.5 (approx)

C) use the same equation and you get
3*92.5 = 0*t + 1/2*10*t^2
t = 7.44 s (approx)

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