5. Projectile Motion

Special Equations in Symmetrical Launches

# A Long Jump on Planet X

Patrick Ford

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Hey, guys, let's check out this problem here involving a long jumper on some unknown planet. So we told some information about their launch speed and the distance they can cover, and we're gonna calculate the gravitational acceleration for this planet. So before I go ahead and draw, you know, and work with any equations, let's just go ahead and draw a quick little sketch of what's going on here. So we have a long jumper who's gonna jump upwards, and then they're gonna return back to the ground again, which is great, because we know we're dealing with the symmetrical launch, so we're gonna be able to use our special equations to solve this problem. So we're told the initial launch speed at this Venus, which is six. And let's just go ahead, draw our paths in the X and y. So this is my ex. This is my Why path. Here, this is a B. And then back down to see again. This is point B. The maximum height. Okay, so what are we actually looking for here? We're looking for the gravitational acceleration on this planet, so that variable is G. So we look at our two special equations here. We've got one for the symmetrical time. And we also have one for the total horizontal displacement or the range. Now what happens is unfortunately, g pops up inside of both of these equations. So remember that both of these equations time and range depend on Venus feta and then G right, so we have enough data and G, it's the same unknowns here. So in order to figure out which one of these equations I'm gonna use, I'm gonna take a look. At what? What? What I know the most information about. So, for example, here we're not told anything about the amount of time that this long jumper spends in the air. However, we do know, we do know that their maximum distance that they can cover turns out to be 9 m. So this whole horizontal distance here, which is our is equal to 9 m. This actually happens to be the maximum horizontal distance covered. So because I know some information about the range, I'm actually just gonna use my range equation. So here's what I'm gonna dio I have are symmetry. So our symmetry is just equal to V, not squared times the sign of two theta divided by G. So I'm looking for here is in looking for the gravitational acceleration. Now I know what the range is. I know what our symmetry is. Basically, this is just the same numbers and I know what the initial launch speed is. All I need to do is I just need to figure out the launch angle and they will be able to solve this equation because I only have one unknown. So what do I What am I told here? How do I figure out data? Well, the other thing I'm told about this range is that the maximum distance possible to cover is equal to 9 m. And so remember that there's a special property about symmetrical launches. The maximum distance that you can cover for any given launch Velocity happens when the data angle is equal to 45 degrees. So that this problem is actually telling us or we can gather from this is that the launch angle is actually 45 degrees. Therefore, we do know what this launch angle is, and so therefore we can solve Fergie. We're just gonna use this equation here and then flip basically trade places between G. And are we just gonna divide you over to the other side and then or multiplied you over to the other side and then divide the are symmetry down. So our equation just becomes G equals V not square times the sign of tooth data two times data divided by our symmetry. So that means that RG ends up being This is gonna be six square times the sign of two times 45 divided by nine. And if you go ahead and work this out, what happens is this just turns into one and this is 36/9, which ends up being 4 m per second squared. So that is our answer. 4 m per second squared is the gravitational acceleration. So therefore we're looking at answer choice B. That's it for this one. Guys, let's keep going

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