# Physics Kinematics of a rock drop

Hey... need some help with this quesiton:

A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 3.2 seconds later. Ifthe speed of sound is 340 m/s, how high is the cliff?

multiplying 3.2 by 340 gets me 1088 m which doesnt seem right.

How do I go about solving this problem?

Thanks much.

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That does not seem right. The question seems rather ambiguous as to whether they mean 3.2 seconds after the rock strikes the ocean or after it is dropped. Try it the other way; 3.2 seconds after the rock is dropped

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H(t) = -4.9t^2+ h0

I do not know h0 so that remains an unknown. IF I add that to v/h then Im gonna have quite a few unknowns.

EnumaElish
Homework Helper
Is h0 the initial speed? If so, shouldn't it be zero?

H0 is imo the initial height.

If h0 = 0 ,

H= (-4.9)(3.2) = 15.68 m

H+ (v/h) = 15.68 + ( 340/15.68) = 37.36 m

Is that correct?

haha are you in my class?

number 47 right Maccully

haha are you in my class?

number 47 right Maccully
huh...........? whats 47? This is from the Giancoli book by Prentice Hall.

its number 47, in the book

probably is.. dunno.... anyhow, still need the answer.. got a test tomorrow on this chapter :(

Time spent falling + time for sound to travel up cliff = 3.2s

To work out the time spent falling reaarange your equation of motion
$$x=ut+\frac{1}{2}at^2$$
$$u=0$$
$$x=\frac{1}{2}at^2$$
$$t=\sqrt{\frac{2x}{a}}$$
The time for the sound to travel up the cliff is x/v, so your equation is:
$$\sqrt{\frac{2x}{a}} + \frac{x}{v} = 3.2$$
which has a solution at 46.02m