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

• B
• Zarich12

#### 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!

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).

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.

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

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

• 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.

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.
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.

• Zarich12
Ok. Thanks!

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).

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}##

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.

I got half the Gs of Svein using 0.005498 seconds to decelerate from 13.86 meters/second to zero in 0.0762 meters.
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!

Thanks to all and good to meet you too.

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

Either way, that's a lot of Gs.

A few comments. It hurts my eyes to see an expression where one input is "about three inches" yield an output with 4 significant digits.

Assuming uniform acceleration gives a lower limit on maximum acceleration. Any non-uniformity will increase the maximum acceleration.

Unless this is in a swamp, 120 g's seems to me to be small. For a 4oz. rocket, that's 30 pounds of force. If you take a similar sized and shaped object, like a railroad spike, it takes substantially more force to push this into typical soil.

• billy_joule