# Rock falls from cliff-Initial Velocity?

• Const@ntine
In summary: Thank you very much for your time and help!In summary, a footballer kicks a rock off a 40.0m cliff with a horizontal velocity. The rock falls into a lake and the footballer hears the "splash" 3.00s later. Sound travels through air with a velocity of 343 m/s. Using the equations Yf = Yi + Viy*t + 1/2*a*t^2 and Xf = Xi + Vix*t, the time that it takes for the rock to fall into the lake is 2.857s. The time it takes for the sound to travel from the lake to the footballer is 0.143s. By using Pythagoras' Theorem and
Const@ntine

## 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.

Last edited:
Darthkostis said:
Sound: R = Vs*t' = 243 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.

This is the correct method. I haven't checked all your working, but it looks close enough to the answer.

When you plug the numbers in immediately, you will generally maximise the rounding error. The only way to get a more accurate answer is to tackle the problem algebraically and plug the numbers in at the end.

PeroK said:
This is the correct method. I haven't checked all your working, but it looks close enough to the answer.

When you plug the numbers in immediately, you will generally maximise the rounding error. The only way to get a more accurate answer is to tackle the problem algebraically and plug the numbers in at the end.

PS The answer I get with ##g = 9.8 ms^{-2}## is ##v = 9.9 ms^{-1}## (to one decimal place).

PeroK said:
This is the correct method. I haven't checked all your working, but it looks close enough to the answer.

When you plug the numbers in immediately, you will generally maximise the rounding error. The only way to get a more accurate answer is to tackle the problem algebraically and plug the numbers in at the end.

I'll try that, and go with the usual methods of rounding errors this time. Thans for the tip!

PeroK said:
PS The answer I get with ##g = 9.8 ms^{-2}## is ##v = 9.9 ms^{-1}## (to one decimal place).

Yeah, that's what I get as well, with slight changes to the 0.0X decimal depending on whether I round the numbers or not.

## 1. What is the initial velocity of a rock falling from a cliff?

The initial velocity of a rock falling from a cliff is the speed at which the rock begins to fall. This velocity is usually determined by the height of the cliff and the force of gravity.

## 2. How is the initial velocity of a rock falling from a cliff calculated?

The initial velocity of a rock falling from a cliff can be calculated using the formula v = √(2gh), where v is the initial velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the cliff.

## 3. Does the initial velocity of a rock falling from a cliff affect its final velocity?

Yes, the initial velocity of a rock falling from a cliff does affect its final velocity. The higher the initial velocity, the faster the rock will be falling when it reaches the ground.

## 4. How does air resistance impact the initial velocity of a rock falling from a cliff?

Air resistance can impact the initial velocity of a rock falling from a cliff by slowing it down. As the rock falls, it will encounter air resistance which will act as a force in the opposite direction of its motion. This force will decrease the rock's initial velocity.

## 5. What factors can affect the initial velocity of a rock falling from a cliff?

The initial velocity of a rock falling from a cliff can be affected by the height of the cliff, the force of gravity, and air resistance. Other factors such as the shape and size of the rock may also play a role in determining the initial velocity.

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