Physics Free fall question, Help Help Help :(?

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Member since: June 13, 2006
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Miechalle
S Physics free fall question help? Thanks :)?
A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 6.1 s later. If the speed of sound is 340 m/s, how high is the cliff? in meters

Answer 1

Let's ignore air resistance, and let's take gravitational acceleration to be 9.8 meters per second squared.

Now let's turn the word problem into an equation. First, the falling rock. We know generally that

d = 1/2 a t^2 ....(1)

where d is distance, a is acceleration, and t is time in seconds.
And since a is 9.8, we get
d = 1/2 9.8 t^2
or
d = 4.9 t^2 ....(2)

where t is the time to fall, in seconds.
Now the sound of the rock reaching our ears:
d = T/340 ....(3)
where d is the same distance, and T is the time in seconds.

Now we also know that
t + T = 6.1 seconds
or
T = 6.1 - t ....(4)
so we can substitute and re-write equation 3 as
d = (6.1-t)/340 ....(5)

Now following the same procedure, we can substitute d from equation (5) into equation (2) like this:
(6.1-t) / 340 = 4.9 t^2

... which gets us down to one equation and one variable. Re-writing using normal algebra gives us
4.9 * 340 t^2 + t - 6.1 = 0

Now use the quadratic formula to solve for t, and then you solve for d using equation (2).

Answer 2

Argh...clearly the rock took some of that 6.1 seconds total elapsed time (T) to fall. That t(f) = sqrt(2h/g) = time to fall. Then the sound travels t(s) = h/v; where v = 340 m/sec. h is the height of the cliff, the answer you are looking for.

Thus, T = t(f) + t(s); so that T^2 = t(f)^2 + 2t(f)t(s) + t(s)^2 and 0 = t(f)^2 + 2t(f)t(s) + t(s)^2 - T^2, which is a very messy quadratic equation. But you can solve it for h because all the other factors are given. [NB: The answer that suggests T^2 = t(f)^2 + t(s)^2 is way off the mark. The times do not form a right triangle, which is what the sum of the squares implies.]

Anyway, the important thing to keep in mind is that there is time elapsed while the rock is falling and then time elapsed while the sound is traveling back up to the thrower.

Answer 3

however long your arm is. He said hes walking along it so his distance form the building never changes. If the question is not a trick question then use your kinematics eqns (you know, 1/2at^2, that stuff) to calc the free fall time. Then you know your profs speed is 1.2 m/s. so in one sec he moves 1.2 meters. You already have the number of seconds the egg was falling from your first eqn so you can figure out the distance he moves in that amount of time. Hope this helped!

Answer 4

speed = distance / time

so, distance = speed * time.

In this case, the sound travels at 340 m/s for 6.1 seconds, so the distance = 340 * 6.1 = 2074m

Light travels about 1 million times faster than sound, so you can ignore the time taken for the light to reach you showing the rock hitting the ocean.

Answer 5

Assuming that the speed of sound stays constant you simply find the height of the cliff by multiplying speed and time

340 m/s * 6.1 s = 2074m

Answer 6

total time = fall time + sound time

from s = 1/2gt^2 (s = height of the cliff)

fall time = sqrt(2*s/g)

sound time = s/sound rate = s/r (r = sound rate)

total time = T = sqrt(2*s/g) + s/r (T = Total time)

T^2 = 2*s/g + s^2/r^2

0 = s^2/r^2 + 2*s/g + T^2
0 = (1/r^2)*s^2 + (2/g)*s + T^2

quadratic equation time
everything is known except for s.

****
i got the height of the cliff as 184m. (actually i got -184 but from the way we set the problem up it is correct since the stone "fell" 184 m. )

Answer 7

340m 1 sec
x m 6.1sec

x=340*6.1
= 2074m