How High Did My Rocket Go?Date: 05/28/99 at 17:46:44 From: Joey Pollino Subject: How high did my rocket go? Dear Dr. Math, We launched a rocket. We want to know how high it went. We were 15 feet away at launch and measured a, an angle for Tangent A, at 89 degrees on the first launch and 67 degrees on the second. I don't have the tables to figure out how high it went. Can you tell me? Joey Date: 05/28/99 at 20:42:58 From: Doctor Peterson Subject: Re: How high did my rocket go? Hi, Joey. My son and I have done a little rocketry too (including the measurement). We can work out the height from your data, and then I'd like to discuss some reasons the answer won't be very accurate. Since you only have one measurement, we have to assume the rocket went straight up. If that is true, it looked like this: ^ /| / | / | / | / |h / | / | /A | +--------+ d The definition of the trigonometric function called the tangent is the ratio of the "opposite" side to the "adjacent" side, so we have tan(A) = h/d Since we know A and d, we can solve for h to get h = d * tan(A) = 15 * tan(89) = 859.35 feet for the first shot = 15 * tan(67) = 35.34 feet for the second I used the calculator on my computer to find the tangents. These are very different numbers! I suspect they are both very inaccurate, for different reasons. First, 89 degrees is so close to vertical that a very small error in your measurement would make a huge difference in height. If it was really 90 degrees (one degree off), that's infinitely high! If it was 88 degrees (one degree in the other direction), it was only 429.54 feet - only half as high as we calculated. Second, 35 feet is so low that I think the rocket may have been blown away from you: ^ //: / / : / / : ./ / : / / :h / / : / / : /A / : +--------+--------+ d If that happens, we no longer have any idea what d is, so our calculation is meaningless. If it blew 15 feet away from you (which could easily happen), it may really be twice as high as we calculated - it looks as if it were at the dot, but it's twice as far and twice as high. The same thing may have happened in reverse in your first flight. It may have blown toward you, so it was straight overhead not because it was so high, but because of where it blew: ^ |\ | \ | \ | \ |h \ | \ | \ |A \ +--------+ d Here the real "d" would be nearly zero, and again you have no idea how high it really is. I should point out that I deliberately gave you a couple of decimal places in my answers above to make it look precise. As you see, decimal places in an answer may be meaningless! So how can you measure more precisely next time? One thing to do is simply to go farther from the rocket: the larger "d" is, the farther A will be from vertical, and the less difference it will make if d is off a little bit because of wind. Even better, there are ways to use two sightings from different places and actually find out where the rocket was in three dimensions - you can know how much it drifted sideways and correct for it. Good books on model rocketry give the formulas for this; since you mentioned a table, you may have seen one. It takes more work to get a double sighting coordinated, but it's really worth it, because you can have a lot more confidence in your results. But just using one sighting and a longer baseline will help. Have fun - I've just given you a good reason to go out and do another launch. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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