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## Homework Statement

A footballer kicks the rock off the cliff, with a horizontal velocity. The cliff's height is 40.0m. The rock falls in a lake. The footballer hears the "splash" 3.00s later. Sound travels through air with a velocity of 343 m/s. What is the rock's initial velocity?

a = -g = -9.80 m/s^2

## Homework Equations

Yf = Yi + Viy*t + 1/2*a*t^2

Xf = Xi + Vix*t

## The Attempt at a Solution

I set the ground as Y = 0, and the footballer's position as X = 0.

Left to Right is (+) and Upwards is (+) as well.

Yf = Yi + Viy*t + 1/2*a*t^2 <=> -40.0 m = -4.90 m/s^2 * t^2 <=> t = 2.857 s

The whole time that it takes for the footballer to kick the rock, for it to fall in the lake and for him to hear the sound, is 3.00s. The rock falls inside the lake in 2.857s. Therefore, the time that the sound takes to travel from the lake to him is t' = 0.143 s.

After that, is where I get stuck. The book's answer is 9.81 m/s, but mine is different. At first, I took sound's velocity as is, going with D = Vs*t' = 343 m/s*0.143s = 49.049 m. And then, I just did D = Vi*t <=> 49.049 m = Vi*2.857s <=> Vi = 17,168 m/s.

Then I thought about it, and figured that it's not specified whether sound moves just vertically, so I took Vsx = Vs*cos(45) = 242.53 m/s and did the same things, but I still don't get the book's result.

I'm probably missing something, but I've been at it for half an hour now, and I'm just not seeing anything else. It's exercise 21, M4 from the 8th Edition of "Physics for Scientists and Engineers".

**EDIT:**I also tried this:

*Sound:*R = Vs*t' = 343 m/s * 0.143s = 49.049 m

So, I've created a triangle, and with Pythagoras' Theorem: D = sqrt(49.049^2-40^2) = 28.386 m

*Rock:*D = Vi*t <=> 28.386m = Vi*2.857s <=> Vi = 9.935 m/s

It's just that I'm not sure how many decimals to keep. I know the laws and rules, but the book sometimes goes with a more round number, and other times it keeps 3 decimals or so.

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