Hey, guys. So in this problem, we're 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 going to be over here. It's going to be served perfectly horizontally. Clear some nets. So I'm going to 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 going to 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 meters away from the player. So we know this distance here is 1.6. And we know that this height 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 going to go ahead and draw the path in the x and y axis. This is my x-axis. It just goes straight from here to here. And the y-axis, if it serves horizontally, then in the y-axis, it's just going to go from here all the way down to this point over here. So what are our points of interest? We have our initial, which is point A, and then nothing else happens, in between, but basically, it's going to hit the net or it's going to go right over the net. So that's going to be our final point, point B. So those are our paths in the x and y. We've got from A to B. So now we're going to figure out what target variable we're looking for. What is that initial speed? What are we actually looking for here? We're looking for v0 (v naught). However, what we do know about horizontal launches is that the initial velocity is purely in the x-axis. So what we're actually looking for is v0, or v0x, but this is also just the same thing as vx. 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 going to be looking for. So now that brings us to the 3rd step. If we're looking for an x variable, the velocity on the x-axis, we're just going to start off with the x-axis equation. There's only one which says that Δx from A to B is equal to vx × t from A to B.
So I'm looking for the initial velocity on the x-axis. So I'm going to need the horizontal displacement, and then I'm going to need the time. So let's take a look at the horizontal displacement. We're told that this ping pong ball is served 1.6 meters away from the net. So that means that this is basically my Δx. So this is my Δx from A to B, and this is equal to 1.6. 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 we're stuck on the x-axis, we're going to have to go to the y-axis. So let's go over here to look for the y axis because I want the time. So we list out all our variables: ay = 9.8, v0y = 0 (because the initial velocity in the y-axis is 0 for horizontal launches), vy is unknown, Δy from A to B = (0.9 - 1.6) = -0.7 meters (falling downwards), and time from A to B. Remember, we're looking for the time here so that we can plug it back into this equation over here. We're looking for the time. That's our target variable.
This is v0x. We know that the initial velocity in the y-axis is 0. What about the final velocity here at point B? We don't know that. What about the vertical displacement from A to B? Well, from A to B, we're actually falling some distance here. But what number are we going to use? We're going to use the 1.6 or the 0.9? Well, it's actually just going to be the difference between those two numbers. Because in our path from A to B, we're actually just falling this distance over here, which if you think about it, is going to be the 0.9 meters that the net is above the ground. So, delta y from A to B is actually negative -0.7. So that gives us 3 out of 5 variables. So this one is ignored, and we pick our equation based on the ignored variables.
So that's actually going to be equation number 3, the one that ignores the final velocity. Delta y from A to B is going to be vaytab+12ayt2. So, now I can just go ahead and fill in everything. I know what delta y is. It's just -0.7. This is going to be -1/2 9.8 and then we've got tAB2. So if you go ahead and solve for this, what you're going to get is that tAB is just equal to 0.25 seconds. So now we just 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 = vx × t from A to B. Now I have both of what those numbers are. And so the final velocity or the initial velocity is going to be the delta x, which is 1.6 divided by 0.25, and I get 6.4 meters per second. And that is the answer. This ping pong ball needs to hit needs to be hit at 6.4 meters per second in order to clear the net on the other side. That's it for this one, guys. That is actually answer choice D. So let me know if you guys have any questions.