Arock is dropped from a sea cliff, and the sound of it striking the ocean in heard 3.0.s later. if the speed of sound is 340m/s, how high is the cliff? the answer is not 1020m.

The height of the cliff is 510 m

The given parameters include;

The height of the cliff is calculated from the principle of echo;

The sound velocity is the ratio of the double distance and time of motion when the sound is heard. This distance is equal to the height of the cliff.

Mathematically, it is written as follows;

Thus, the height of the cliff is 510 m

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s = 51.69 m

Explanation:

For a falling object, an equation of motion is  

s = ut + (1/2)gt2  where,  

s = distance dropped  

u = initial velocity

t = the time of fall

g = acceln due to gravity

Since the rock is dropped, rather than thrown, from the top of the cliff, then its initial velocity, u, is zero. So we can write,

s = (1/2)gt2

or,

s = 4.9t2 (g = 9.8 m/s2)

Let t be the time for the rock to fall the height of the cliff. We also know that the time for the rock to reach the bottom of the cliff, and for the sound of it hitting the water to then reach the top of the cliff is 3.4 secs. That means that the time for the sound to travel up the height of the cliff is (3.4 - t) secs.

Since the speed of sound is 340 m/s, then we can write,

s = (3.4 - t)*340

Since s is the same in both equations, then we can equate them to give,

(3.4 - t)*340 = 4.9t2

Rearranging,

4.9t2 + 340t - 1156 = 0

Using the quadratic formula to solve this quadratic equation,

t = {-340 + sqrt[3402 - 4*4.9*(-1156)]}/(2*4.9)  

t = {-340 + 371.83}/9.8

t = 3.248 s

Using either of the two original equations for s,  

s = 4.9*(3.248)2

s = 51.69 m

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

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from wikipedia