*NOTE At end of video I say trajectory does not change, of course it does becoming vertical with height of falling. But this is the same for both masses and they land at the same time but the 2v with 4x the distance.
For 7:11, this isn't true. The horizontal distance the 2x velocity object will travel will depend on the times they spend in air. If they are launched horizontally, then they spend the same amount of time in air before dropping (ignoring air resistance, etc.), so the 2v object will travel 2x distance. When the initial launch angle changes, the times both will spend in air will not be equal, but it's going to vary. In fact, I'm pretty sure when you have an initial launch angle upwards, the 2v object will travel NEITHER 2x distance nor 4x distance but something in-between that will vary with the launch angle. We'd have to do the math directly, but I'm not sure if it is worth it. If you wish, you can request me to do the math precisely.
@@Ian.Gostling The kinetic energy of the bob is converted into the kinetic energy of the block (½mv²), then the kinetic energy of the block is *linearly* dissipated into heat by friction, and that means that twice the distance equals twice the energy, yes. Perfect experiment, perfect video, Ian. 😃👍
@@_John_Sean_Walker Thanks John, that's kind but I know it's not perfect, but only another good indication we have not been hoodwinked by the pseudo religious scientific fabalists.
Nice demonstration Ian, please don't let Goofy get to you mate, the ¢unt feeds on it. I agree with John, the block going askew will mean all the energy didn't go into the block, but nice clear demo of the energy despite what horse $hit Goofy will spout. Don't expect an apology from the goon either mate, as when your perfect in every conceivable way, there is nothing to apologise for, obviously. 😂🤣👍
@@Ian.Gostling He is full of $hit, once the balls are sepeated by 4x the distance, and no longer moving horizontally, they will only move in the downward direction, however far they fall, so the distance between them will remain the same There is a menegarie of elephants in the room, that the coward can't even attempt to explain, he lives in his own tiny mind.👍😊
@@Ian.GostlingI may stand corrected on my last commemt regarding the 4x distance, MM left a reply regarding it and he certainly knows his onions, so I may look into it more, but MM suggests somewhere in between 2 and 4 distance, depending on launch angle, which makes sense as that would alter the fall time, physics is never as straight forward as we first think. In regards to commemtimg on his channel mate, people do, but he just blocks everyone or deletes the comments, he is not an honest player. Video responses are ignored too, he doesn't want discussion or debate, he wants to belittle folk to feel better about himself and just seeks validation and adoration for his crackpot ideas, he will be waiting a very long time for that. 😂👍
@@michaelfowell223 Since the second probe was dropped from a higher point, we may assume that it takes the second probe a longer time to reach the point of the actual launching into the air, and in addition, we may assume that it also has a faster speed, so for these reasons the second probe will be a longer time in the air before it hits the carpet, right? And when the second probe has a faster speed and stays longer in the air, it therefore must be traveling a larger distance before it falls on the carpet too, right? And when the second probe has twice the velocity as the first probe has, then the second probe will stay twice as long in the air, and it will therefore travel 2 × 2 = 4 times further. The 45° angle gives the largest difference in distance traveled because a smaller angle gives a shorter time in the air, and a larger angle gives a slower forward velocity. Nota bene: The equation for distance traveled when at constant velocity: s = vt
