A little too much estimation…
I’m a big fan of either using 10 m/s/s or 22 mph per second for g, and giving students reasonable numbers when they are doing exercises that involve projectile motion. The real world won’t be so kind when they take data, so I want to start them off where they have a fighting change of reasoning proportionally on the basis of concepts. I’m trying to fight the “iterative plug-n-chug” approach, by doing things we can think through.
But, in one of my classes, I’ll need to be using 9.8 m/s/s, because that’s what they’ll have on their exam. And I won’t have reasonable numbers always, because the problems I’m supposed to provide example are not picked by me. So, I have decided to do a lot of “figure out a number you know is too high”, “figure out a number that is too low”, and “make a good guess”, before doing any plugging into equations.
Take for example, this kind of question I am supposed to work through as an example:
A ball is thrown vertically upward with a speed of 7.2 m/s. On the way down the ball hits a flag pole that’s 1.2 meter above from where it was thrown.
It’s a goalless problem, so I’ll have to decide something to solve for. Let’s go with, “How long will the ball take to get to its highest point?” … Here we go. Well, we know that the ball loses 9.8 m/s every second, but it only needs to lose 7.2 m/s. That must mean that the time to reach its highest point is less than 1 second. How much less? Well, it’s certainly greater than half of a second, because half a second would be time to lose 4.9 m/s. So, a number that is too high is 1 second, and a number that is too low is 0.5 s. I would guess it’s probably pretty close to 3/4 of second, because 7.2 seems pretty close to in between 4.9 and 9.8.
What’s the actual answer? About .73 seconds.
OK. Now on to another question. “How high will it go?” Well, the average velocity of the ball during the trip will be right in the middle of its starting velocity (7.2 m/s) and its ending velocity (0 m/s). That makes its average velocity 3.6 m/s–I can divide by two in my head. Given that I’ve concluded in the first part a travel time of 3/4 of a second, this must mean it must have traveled less than 3.6 m, because it would have taken an entire second to travel 3.6 m, and it didn’t have that much time. My reasoning is that it should go 3/4 of the way to 3.6m in 3/4 of a second. This is actually pretty easy to reason about, because 36 = 9 * 4. So, I’d say it would have gone about 2.7 meters during that time.
OK, so what’s the actual answer? 2.6 meters
OK, so how much time to hit the flag? Well, it took 3/4 of a second to go up. It would have taken another 3/4 of a second to go down, but it didn’t quite go all the way down because it hit the flag. So, How far did it get? Well, it got a little over half way down (2.7-1.2 = 1.5). Let’s just call it half for good measure. Now it might be tempting to think it took half the time to go half way down as the whole way down. That would be true if it was moving at constant velocity. But it’s moving slower earlier and faster later, because it takes time to speed up. Given that, we know that it must take MORE than half the time to cover half the distance, because that’s the slow part of the motion. By that alone we can say that it must have taken more than 3/4 + 3/8 = 9/8 of a second. So we have a lower bound of 1.1 s and an upper bound of 1.5 seconds. Let’s just guess that’s it took 0.5 second, because that’s more than half of .75 seconds.
That would mean the whole trip should be about 1.25 s = 0.75s on the way + 0.5s on the way down seconds.
The real answer? 1.2 seconds.
OK. So, some of this is a bit much. But, I didn’t have to use equations.