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Clearing the Net

Patrick Ford
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Hey, guys. So this problem We were trying to figure out the initial speed that a Ping Pong ball needs to have in order to clear a net on the table. We know that this Ping pong ball is served horizontally, so let's go ahead and just draw a quick diagram of what's going on. So we have a ball that's gonna be over here, it's gonna be served perfectly horizontally. We know this ball is served 1.2 m off of the ground, and what it needs to do is some distance later. It needs to clear some nets. So I'm gonna draw this little point right here. So what happens is this ball needs to have some initial velocity so that when it follows its parabolic path, it's gonna just barely go over this net right here. So that's what this sort of path looks like right here. We're told also that this net is 1.6 m away from the player. So this is we know this distance here is 1.6 and we know this heights here of the net off of the table is 0.9. Alright, so that's what we know about this problem. So now we're just gonna go ahead and draw the past in the X and Y We're just gonna stick to the steps, right? So we're gonna draw the paths in the X and Y axis. So this is my ex access. It just goes from straight from here to here and the Y axis. If it served horizontally, then in the Y axis is just gonna go from here all the way down to this point over here. So what are points of interest? We have our initial, which is point a, and then nothing else happens in between. But basically, it's gonna hit the Nets, or it's gonna go right over the nets. So that's gonna be our final point. Point B. So those are our paths in the X and Y. We've got from A to B. And so now we're going to figure out what target variable we're looking for. What is that initial speed? What we actually looking for here? What? We're looking for V knots, right? So this needs to have some initial velocity V knots. However, what we do know from initial velocities is that we do know about horizontal launches is that the initial velocity is purely in the X axis. So we're actually looking for is V not or Vienna X, But this is also just the same thing as Vieques. It's basically just the same horizontal velocity throughout the entire motion. So all these three variables really mean the same thing? So that's really what we're gonna be looking for. So now that brings us to the third step. What interval or we're gonna look at We're looking for the initial velocity and we're looking basically just between the time that it served and the point where it hits the net or goes over the net, Then we're just gonna look at the interval from A to B. Okay, so we're looking for a X variable The velocity in the X axis We're just going to start off with the X axis equation. There's only one which says that Delta X from A to B is equal to v x times t from A to B. So this is what I'm looking for. I'm looking for the initial velocity in the X axis, so I'm gonna need the horizontal displacement and I'm gonna need the time So let's take a look at the horizontal displacement. We were told that this Ping pong ball is served 1.6 m away from the net. So that means that this is basically my Delta X. So this is my delta X from A to B, and this is equal to 1.6. So I'm just actually gonna erase this over here because it's actually Mawr. Ah, sort of. You could see it better here that it's 1.6. All right, so we know what this number is. Unfortunately, we don't know how long it takes to go from A to B, so we actually don't know how long that is. So we're a little bit stuck with this equation. We can't use it. So therefore, when wherever we're stuck in the x axis, we're gonna have to go to the Y axis. So we go over here to look for the y axis because I want time. So I'm gonna listen at all my variables. A y equals 9.8 v. Not y is v a Y v y is v b y. We've got Delta y from A to B and we've got time from a to B. Remember, we're looking for the time here, so we can actually plug it back into this equation over here. So we're looking for the time. That's our target variable. So we got we need three out of five variables. So what about the initial velocity in the Y Axis V A. Y? Well, remember, what's special about horizontal launch problems is that your V a y. Your initial velocity in the Y axis is just equal to zero. And so all of your initial velocity is just purely in the X axis. And that's just the speed that was given to us. Or actually, in this case, the speed that we're gonna look for this is V X. All right, so we know that the initial velocity in the Y Axis zero What about the final velocity here at point B? So what is the components of the velocity V B Y? We actually don't know what that is. So this is gonna be We don't know what that is. What about the delta y from A to B? What about the horizontal? Sorry, The vertical displacement from A to B Well, from from A to B. We're actually falling some distance here, but what number we're gonna use are we gonna use the 1.2 or we're gonna use the 0.9. Well, it's actually just gonna be the difference between those two numbers, because in our path, from A to B or actually just falling this distance over here, which, if you think about it, is gonna be the 1.2 m that we're serving off of the ground to 0. m that the net is above the ground. So it's whatever the difference of those two numbers is. So it means that Delta Y from A to B is actually negative 0.3. It's negative because this displacement points downwards. So this is gonna be negative 0.3, and so that gives us three out of five variables. So this one is the ignored, and we pick our equation based on the ignored variables. That's actually gonna be equation number three, the one that ignores the final velocity. So Delta y from A to B is going to be, um, let's see. I've got v a Y t a b plus one half a Y t a B squared. Alright, so it's again. It's also useful about horizontal launch equations. Or what's convenient is that this term will always go away. Because remember that V A Y is equal to zero. So the only just leaves us with two terms. So now I could just go ahead and fill in everything. I know what Delta Y is. It's just negative. 0.3, this is gonna be negative. One half negative, 9.8, that's G. And then you got ta be squared. So if you go ahead and soul for this, what you're gonna get is the TA be is just equal to 0. seconds. So now we're just gonna plug this back into our X axis equation and then we'll be able to figure out that horizontal velocity So my delta X from A to B equals v x times ta to be Now I have both with those numbers are and so the final velocity or the initial velocity is going to be the Delta X, which is the 1.6 divided by 0.25. I get 6.4 m per second and that is the answer this Ping pong ball needs to hit. Needs to be hit at 6.4 m per second in order to clear the net on the other side. That's it. But this one, guys, that actually is answer choice D. So let me know if you guys have any questions.