# Couple of kinematics problems

1. Apr 16, 2004

### Phantom8295

Can any one suggest how to do these problems.

1) A rocks is dropped from a sea cliff and the sound of it striking the ocean is heard 3.4s later. If the speed of sound is 340m/s, how high is the cliff?

2) An apple is dropped and a multi-flash camera is used to take photographs of the falling apple. If the apple is of 10c in diameter, estimate the time between each photoflash of the apple?

I got really stuck with these two problems. Any suggestions how to solve them?

Thanks
P.

2. Apr 16, 2004

### holly

Over on page 3 is a similar question entitled "Simple Problem: Depth of Well providing sound reaches you in 1.5 seconds" and Doc Al (one of the bigger of the Brains on PF) said how to do it, something about quadratics. Maybe you'll get lucky and he'll answer it again...

That second problem, I don't know at all. Guess you'll need to find its volume (it's sort of a sphere) and think of it falling down at the gravitational constant. Guess they want to have a clear shot of it each time?

Good luck, hope one of the Brains answers you.

3. Apr 16, 2004

### Chen

Let H be the height of the cliff. The function describing the rock's location is:

$$y_{(t)} = \frac{1}{2}gt^2$$

From there you can find the time it takes the rock to reach the bottom of the cliff:

$$y_{(t = t_f)} = \frac{1}{2}gt_f^2 = H$$

$$t_f = \sqrt{\frac{2H}{g}}$$

Now, how much time does it take sound to travel a distance of H (from the bottom of the cliff to your ears) at a speed of 340m/s?

$$H = v_st_s$$

$$t_s = \frac{H}{v_s}$$

Finally, you know that the sum of these two time intervals should equal 3.4 seconds:

$$t_f + t_s = 3.4$$

$$\sqrt{\frac{2H}{g}} + \frac{H}{v_s} = 3.4$$

If you let $$x = \sqrt{H}$$ you get this equation:

$$\frac{x^2}{v_s} + x\sqrt{\frac{2}{g}} - 3.4 = 0$$

Now solve for x and then find H. I get H = 51.7 meters.

Last edited: Apr 16, 2004
4. Apr 16, 2004

### Phantom8295

thanks chen

I solved that problem.

As for the apple problem, its basically an apple is dropped and the photos are taken at regular intervals. The shot does not have to be a complete apple, as it falls down the acceleration increases and the distance between the previous position to current position increases at each shot. The questions is how many shots can be taken in a sec, or whats the time interval between each shot. Hope i am making it clear. I am still clueless about it. Have to try it again tomorrow. Have a good night folks.

P.

5. Apr 16, 2004

That really depends on what kind of camera you're using, doesn't it?

You could program the camera to take pictures such that the distance between each image of the apple is the same, or you could take the pictures at a constant rate to see the acceleration.

6. Apr 16, 2004

### holly

They must want a clear shot of the apple each time. That's why they give a hint on its size. Since it's going to fall approx 5 m during the second, how many lengths of apple is that? That may be what they want.

7. Apr 16, 2004

They want him to find the distance between r(0s) and r(1s)? Seems kinda easy...

8. Apr 17, 2004

### holly

E A S Y ? ? ?

Easy for whom? Yeah, for someone with a big ol' brain, maybe...

9. Apr 17, 2004

### jdavel

1) A rocks is dropped from a sea cliff and the sound of it striking the ocean is heard 3.4s later. If the speed of sound is 340m/s, how high is the cliff?

Edit: Well I see Chen did your homework for you, so you don't need my help below. But you didn't learn anything either; nice work Chen!

The time interval from when the rock is dropped to when the sound is heard can be divided into two parts (t1 and t2) that add to give 3.4secs. What does each part mean? Can you find an equation for each one? Is there a variable that represents the height of the cliff in each one?

2) An apple is dropped and a multi-flash camera is used to take photographs of the falling apple. If the apple is of 10c in diameter, estimate the time between each photoflash of the apple?

Check the original wording of the problem; you're leaving something out.

Last edited: Apr 17, 2004
10. Apr 17, 2004