# B What is the Gs of an object hitting the ground at 31 mph?

1. Apr 23, 2016

### Zarich12

I launched a model rocket, and like a dummy, put the nose cone on too tight. From its apogee (aproximately 1000 ft.) it fell to Earth at a terminal velocity of 31 mph. It ran into firmly packed dirt and no rocks. I want to know how many Gs the rocket was under when it hit the ground and if it's worth looking for. Thanks!

2. Apr 23, 2016

### Svein

You must tell us how deeply into the ground it ended up - then use the standard formulas (s = v0t and s = 0.5 at2) gives a = 2v02/s).

3. Apr 23, 2016

### Zarich12

It is probably embeded 3 inches into the ground. Sorry, but I don't know how to use this formula as I'm 13. If you have an estimate that's all I need.

4. Apr 23, 2016

### Zarich12

Or if you could give what all the variables stand for I could do the math.

5. Apr 23, 2016

### Svein

Since I am from a metric country, I have to translate:
• 31mph = 13.86m/s
• 3inches = 0.0762m
• ∴ a = 5042m/s2
• g ≈ 9.81m/s2, so approximately 514Gs

6. Apr 23, 2016

### Zarich12

Thanks! One more question. If the that same rocket were made of cardboard and balsa wood fins, would it be worth getting back or would it be totaly destroyed.

7. Apr 23, 2016

### Svein

My calculations were for a solid (a steel ball, a rock or something similar). Cardboard would deform and probably be destroyed, but thereby creating a longer retardation distance for objects inside, thereby sort of protecting them.

8. Apr 23, 2016

### Zarich12

Ok. Thanks!

9. Apr 23, 2016

### jbriggs444

Try $a=\frac{v^2}{2s}$ instead.

If you do not want to Google SUVAT, you can arrive at it in a couple of ways. One way is with the work energy theorem.

$Fs=\frac{1}{2}mv^2$

Substitute for $F$ using $F=ma$

$mas=\frac{1}{2}mv^2$

Divide by m and s giving the expected result:

$a=\frac{v^2}{2s}$

Another way is to compute the average velocity during the acceleration as half of the impact velocity.

$v_{avg}=\frac{v}{2}$

Impact duration is distance divided by the average velocity

$t=\frac{s}{\frac{v}{2}} = \frac{2s}{v}$

Acceleration is impact velocity divided by duration giving the expected result:

$a=\frac{v}{\frac{2s}{v}} = \frac{v^2}{2s}$

10. Apr 24, 2016

### Barbarian_Geek

I got half the Gs of Svein using 0.005498 seconds to decelerate from 13.86 meters/second to zero in 0.0762 meters. Using the basic formula: acceleration = velocity ÷ time, I got 2521 meters/second2, or 257Gs.

11. Apr 24, 2016

### jbriggs444

That figure of 0.005498 results from dividing 0.0762 meters by 13.86 meters/second. $t=\frac{s}{v}$. But that assumes that the average velocity during the deceleration is equal to the initial impact velocity. That assumption is obviously wrong. A more reasonable assumption is constant deceleration, in which case the relevant velocity is the average of the initial impact velocity and the final velocity (rest). That is 6.93 meters per second (half of 13.86 m/s).

That's the second derivation in #9 above.

You get half the G's of Svein because Svein's result was off by a factor of four. Your result is only off by a factor of two.

Almost forgot... Welcome to Physics Forums. Good to meet you!

12. Apr 24, 2016

### Zarich12

Thanks to all and good to meet you too.

13. Apr 24, 2016

### Biker

Weird result.

I have used the same ways that Jbriggs444 used. It is different than Svein's answer.

So for the sake of not duplicated the same thing just use the final equation that Jbriggs444 showed.

a = v^2 / 2s Where v is the impact velocity and s is the distance
a = 13.85^2 / (2 * 0.0762) = 1258.6

Dividing it by g which is 9.8 m/s^2 gives you 128.4g

14. Apr 24, 2016

### Zarich12

Either way, that's a lot of Gs.

15. Apr 24, 2016