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Anderson Video - Time in Air During a Dunk

Professor Anderson
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 >> Hello class, Professor Anderson here. Let's look at a classic physics problem which is a basketball player jumping up to dunk a basketball. And this is a question that a lot of physics researchers have asked various ages of kids to see what they think. All right, so let’s— I was going to say basketball player Michael Jordan, but you guys are all too young, you probably don't even know who Michael Jordan is anymore, right? So we've got to get a little more current. So then I was thinking, well who's the next most famous basketball player. Well, Chris Dudley probably is pretty famous, right? Everybody remembers him, no? Let's go to a baseball player that you do know, SDSU basketball player Xavier Thames. So if he jumps up to dunk a basketball, how long is he in the air? And let's give you a couple of ranges here. Let's say that the total time could be a half a second, one second, two seconds, five seconds, ten seconds, and let's take a little vote amongst the people here, okay? How many people think that it is going to be ten seconds, just a show of hands? Nobody likes ten seconds, okay. What about five seconds? No, nobody likes five seconds. What about two seconds? Two seconds sound reasonable? Okay. Two people said two seconds. What about one second? One second, okay? Five people like that. And then what about a half a second? All right, three people like that. So, we've got a nice Gaussian distribution there between choices and it looks like the average of all those is somewhere around one second. So that's our guess. We're going to say one second that he's in the air. How do we figure this out? Well, first off we need to know a little bit about basketball. How high is the basketball hoop off the ground? >> Ten feet. >> Ten feet, right? Regulation basketball hoop is ten feet off the ground. So does he have to jump ten feet in the air? >> No. >> No. He doesn't have to jump ten feet in the air because he's pretty tall and he's got long arms, right? Maybe not that long, okay. But he's already pretty close to the rim when he does that, so he doesn't have to jump ten feet. He only has to jump this far, from there to there, and that is a considerably smaller amount to jump. So basketball players have good vertical leap which they can go pretty high, but pretty high for a human is on the order of three feet, okay? So let's say that h is one meter, that's how far he has to go. One meter and we're approximating this. And let's now calculate how long he is in the air. So we are not given a whole lot of information here, right? We don't have anything about the initial velocity as he jumps off the ground, and somehow we've got to come up with this time. So how can we attack this problem? So he said that the motion is in fact symmetric in this case. Going up is exactly the same as going down. He's going to take the same amount of time to go up as he takes to go down, and so let's just concentrate on one half of the motion. Let's look at the time it would take for somebody to fall a distance of one meter. [ Background Sounds ] Okay. So let's take a look at how long it would take for someone to fall a distance of one meter. Okay. That's not too bad. What can we say about this? Well, a person falling one meter is no different than any other object falling one meter. So we can just consider a rock that we drop, and if we want to describe that motion, what sort of equation should we go back to right away? >> Position. The kinematic equations for >> Position. Position. Absolutely. One of those looks like this: Yf equals Yi plus Vyi times t plus one half A sub y t squared. That looks like a pretty good equation to use. Do we know y final? >> Yes. Yes. That is zero. Do we know y initial? >> Yes. Yes. That is the height h. What about Vy initial? What's Vy initial? Zero. What is ay? >> g. Negative g. So we put a minus sign out in front and we're going to label this one half gt squared right there, and now look. We have one equation and the only unknown in there is t, and so we can solve this very quickly for t. I get one half gt squared equals h. I just move that over to the other side and now I multiply by two so I get gt squared equals 2h. And then I divide by g and I get t equals 2h over g and I take the square root, okay? So we can plug in these numbers. If we do that, what do we get? We get two times one meter divided by 9.8. You guys, somebody punch into your calculator those numbers and tell me what you get for your answer. [ Background Sounds ] >> .45 0.45 seconds, okay? So that is the time to fall and so the total time is of course just twice that which is 0.9 seconds which is pretty close to our guess of one second, okay. We approximated the height that he can jump to but this is a very good guess. Now there's something a little interesting about our answer down here which is the following. When we did the square root, what do you get when you take a square root? >> Plus or minus. Exactly. Plus or minus. So really we should put plus or minus here. And now you say, well wait a minute. What is the significance of the minus, right? We just assumed that it was positive .45 seconds, but there's a minus there. Does that have any physical significance? We set the clock to t equals zero up at the top, but if I run that clock backwards in time, that minus .45 exactly corresponds to how long it took them to go up, okay? So the fact that the motion is symmetric means that the clock can run forward and we get .45, or it can run backwards and we get the negative .45, okay. So that actually does have physical significance which is kind of cool. All right. Hopefully that's clear, if not, come see me in my office hours. Cheers.