Steel Drum From

what is th speed of slingshot?

To find the speed of a ball bearing launched from a wrist rocket (sling-shot) you take aim at a 55 gallon steel drum exactly 80 meters from you. You hear the "clunk" of impact 0.68 seconds after you "let fly." If the speed of sound on that particular day is 325 m/s, how fast must the ball bearing be traveling?

You release the bearing at time 0. The bearing travels at speed v for some time t1. Total distance travelled: v*t1 = 80m. The bearing hits the drum at time t1. CLUNK!

The CLUNK! sound starts traveling at time t1,and travels at velocity 325 m/s for time t2. Total distance is still 80m, so (325 m/s)*t2=80m. You hear the CLUNK!

Total elapsed time is t1+t2 = 0.68s.

Now we have a system of equations. Do not forget to write down and keep track of the units!

v * t1 = 80m
(325 m/s) * t2 = 80m
t1 + t2 = 0.68s

We have three equations in three unknowns (t1, t2, and v). We want to know v. Note that all three must be positive, so we know that, too, even though it really doesn't matter here... But it might help you catch an error.

We can get t2 easily.

(325 m/s) * t2 = 80m
t2 = 80m / (325 m/s) = 80/325 s = 16/65 s

Now we can solve for t1.

t1 + t2 = 0.68s
t1 + (16/65 s) = 0.68s
t1 = 0.68 s - 16/65 s = 0.43 s

Note that we used two significant digits there because we only now two digits of the first number. That is, we know 0.68, but not more, like 0.68054, so it is wrong to write down more digits.

Now that we know t1, we can solve for v, and get the answer.

v * t1 = 80m
v * (0.43s) = 80m
v = (80m) / (0.43s)
v = 190 m/s

Again, we only know two digits of each number, so we just round to two significant digits in our answer.

How fast is that? Well, in kph we get the following.

(190 m/s) * (1 km / 1000 m) * (3600 s / 1 hour)
= (190) * (3600 / 1000) kph
= (190) * (18/5) kph
= (190/5) * 18 kph
= 38 * 18 kph
= 684 kph

Or 680 kph to two significant digits. That's roughly 420 mph.

Note that 1km = 1000m, so when we multiply by (1km / 1000m) we are really just multiplying by one. We're always allowed to do that. Same for (3600s / 1 hour).

Does that answer make sense? Let's do a quick sanity check.

The bearing is traveling at 190 m/s, and the speed of sound is 325 m/s. The ratios of the speeds is 190 / 325. The time for the flight of the bearing is 0.43 s, and the time for the pressure wave to reach you from the drum is 0.25 s. The ratio of the two is 0.25 / 0.43. Since speed and time are directly, and linearly, related, these ratios should be the same. They are (to two significant digits), so we conclude that our answer is correct.

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This entry was posted on Saturday, June 19th, 2010 at 11:14 am and is filed under Instruments. You can follow any responses to this entry through the RSS 2.0 feed. Both comments and pings are currently closed.