Question

I REALLY NEED TO LEARN THIS PROBLEM INSIDE OUT. PLEASE BE AS DERCRIPTIVE AND CLEAR AS YOU CAN BE PLEASE ( AS IF YOU WERE TEACHING A PERSON WHO KNOWS NOTHING ABOUT PHYSICS) thank so much

1. a rock is dropping from a sea cliff and the sound of it striking the ocean is heard 3.4 s later. if the speed of sound is 340 m/s, how high is the cliff?

Answer 1

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