Professor Anderson

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>> Ok let's drop an object straight down now and let's see if we can calculate the speed just before it hits the ground. Ok so here we are on the earth and we're going to take an object and we will let it go from height h and it's going to fall towards the earth and we want to know what this final speed is. Well we can do this again with the kinematic equations or we can just go straight back to conservation of energy. And conservation of energy is really very straight forward in this case. Initial equals final, what do we have initially? We have some potential energy, we have some kinetic energy, on the final side we have potential energy, we have some kinetic energy. And if we starting from -- if we're starting up at a height h, then the potential is just mgh. If we're starting at a speed of zero, add a zero. If we are ending at the ground, final height is zero. And if we're moving just before we hit the ground, we've got kinetic energy of 1/2mv squared. And now this should look very familiar, right, this looks exactly like we did with the skier on the slope. We get vf is equal to the square root of 2gh again. Pretty cool and pretty straight forward I think. Very simple derivation. But let's take this problem now and let's slightly modify it. Let's throw the object up. Now let's calculate what the vf is when it comes back down. So instead of dropping this object straight down, we're going to do this. We're going to throw it up at initial speed vi, it's going to go up for a little bit, and then it's going to come back down, and it's going to hit the ground at vf. And let's see if we can figure out what that vf is. Ok, energy initial equals energy final. We're up here at a height, h, so we've got potential energy of mgh. We threw it at speed vi and so we have initial kinetic energy of 1/2mvi squared. When it comes back down we don't have any more potential energy because we're down at the ground. But we do have kinetic energy. Ok, and now we can solve this thing for vf. How do we do it? Well, we can cross out the m's on both sides, if I multiply everything by two then the 1/2's go away and I put a two out in front of that thing. And so we get vf is equal to 2gh plus vi squared and I'm going to take the square root of that whole thing. And now you look at this result and you want to make sense of it with what we just did where we dropped the object. That would be the case where vi equals zero. And so if vi equals zero then vf is just square root of 2gh, like we found before. If we increase vi then it's going to hit the ground even faster. But there's something sort of interesting here right. The interesting thing is we didn't tell you what direction you threw this object. So we said you're going to throw it up, but did we say you're going to throw it straight up? Not necessarily. And in fact, let's say we didn't throw it up; we threw it out at an angle. If we threw this object out at this angle with speed vi. Comes down here, hits the ground at vf, and let's say that is angle theta. What is this vf? I'm asking you guys. Naseem, can you hand the mic to Naseem? Naseem here I am, I'm standing on a building at height, h, I throw this object at vi out at this angle. It's going to hit the ground at vf, what is that speed, vf? >> (student speaking) Would it be the square root of 2gh plus vi squared? >> Good guess. Where did you come up with that? It's the exact same thing. Ok, you know that because if you look at this equation, we could write down the exact same equation for this problem. What's the potential energy? Mgh. What's the kinetic energy initially? 1/2 mvi squared. What's the final energy? 1/2mv squared. It's the exact same equation. And that's because it's a scalar equation. There's no direction associated with conservation of energy. There's no direction associated with energy. It's a number. It's 10 Joules, or 20 Joules, it is a number. And so it's actually independent of this angle, theta. It doesn't matter what direction you throw this thing. It's always going to hit the ground at that speed vf. All it depends on is how high you start it from and how fast you threw it, and that's it. Which I think is kind of cool. Ok, nice simplifying feature of conservation of energy. Alright any questions about this before we move on to something else? Everybody ok with this? Alright.

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>> Ok let's drop an object straight down now and let's see if we can calculate the speed just before it hits the ground. Ok so here we are on the earth and we're going to take an object and we will let it go from height h and it's going to fall towards the earth and we want to know what this final speed is. Well we can do this again with the kinematic equations or we can just go straight back to conservation of energy. And conservation of energy is really very straight forward in this case. Initial equals final, what do we have initially? We have some potential energy, we have some kinetic energy, on the final side we have potential energy, we have some kinetic energy. And if we starting from -- if we're starting up at a height h, then the potential is just mgh. If we're starting at a speed of zero, add a zero. If we are ending at the ground, final height is zero. And if we're moving just before we hit the ground, we've got kinetic energy of 1/2mv squared. And now this should look very familiar, right, this looks exactly like we did with the skier on the slope. We get vf is equal to the square root of 2gh again. Pretty cool and pretty straight forward I think. Very simple derivation. But let's take this problem now and let's slightly modify it. Let's throw the object up. Now let's calculate what the vf is when it comes back down. So instead of dropping this object straight down, we're going to do this. We're going to throw it up at initial speed vi, it's going to go up for a little bit, and then it's going to come back down, and it's going to hit the ground at vf. And let's see if we can figure out what that vf is. Ok, energy initial equals energy final. We're up here at a height, h, so we've got potential energy of mgh. We threw it at speed vi and so we have initial kinetic energy of 1/2mvi squared. When it comes back down we don't have any more potential energy because we're down at the ground. But we do have kinetic energy. Ok, and now we can solve this thing for vf. How do we do it? Well, we can cross out the m's on both sides, if I multiply everything by two then the 1/2's go away and I put a two out in front of that thing. And so we get vf is equal to 2gh plus vi squared and I'm going to take the square root of that whole thing. And now you look at this result and you want to make sense of it with what we just did where we dropped the object. That would be the case where vi equals zero. And so if vi equals zero then vf is just square root of 2gh, like we found before. If we increase vi then it's going to hit the ground even faster. But there's something sort of interesting here right. The interesting thing is we didn't tell you what direction you threw this object. So we said you're going to throw it up, but did we say you're going to throw it straight up? Not necessarily. And in fact, let's say we didn't throw it up; we threw it out at an angle. If we threw this object out at this angle with speed vi. Comes down here, hits the ground at vf, and let's say that is angle theta. What is this vf? I'm asking you guys. Naseem, can you hand the mic to Naseem? Naseem here I am, I'm standing on a building at height, h, I throw this object at vi out at this angle. It's going to hit the ground at vf, what is that speed, vf? >> (student speaking) Would it be the square root of 2gh plus vi squared? >> Good guess. Where did you come up with that? It's the exact same thing. Ok, you know that because if you look at this equation, we could write down the exact same equation for this problem. What's the potential energy? Mgh. What's the kinetic energy initially? 1/2 mvi squared. What's the final energy? 1/2mv squared. It's the exact same equation. And that's because it's a scalar equation. There's no direction associated with conservation of energy. There's no direction associated with energy. It's a number. It's 10 Joules, or 20 Joules, it is a number. And so it's actually independent of this angle, theta. It doesn't matter what direction you throw this thing. It's always going to hit the ground at that speed vf. All it depends on is how high you start it from and how fast you threw it, and that's it. Which I think is kind of cool. Ok, nice simplifying feature of conservation of energy. Alright any questions about this before we move on to something else? Everybody ok with this? Alright.