Hey, guys. So let's go ahead and check out this problem here. You're going to launch a 4 kilogram object directly up from the ground. So I go to the ground like this. You're going to take this 4 kilogram ball and you're going to throw it up with some initial speed \( v_0 = 40 \). And we want to figure out basically what happens when the ball finally comes to a stop here at some maximum height. So here, if you take the ground level to be 0, what we want to calculate is what is this \( y_{\text{max}} \) value here. Alright? So we've got some changing speeds and changing heights. We know we're going to use energy conservation. So we've already got the diagram and we're going to write our energy conservation equation. So we're going to write \( K_{\text{initial}} + U_{\text{initial}} = K_{\text{final}} + U_{\text{final}} \). Now we're going to eliminate and expand out each one of our terms here. Alright? We've got some initial kinetic energy. That's the initial speed, of 40 meters per second, so we've got that. But here, when we're at the ground level, if we take \( y = 0 \) to be the ground, right, where we're starting from, then that means that our gravitational potential energy is 0 here, so, therefore, there is no gravitational potential. Alright. So there's nothing there. What about \( K_{\text{final}} \)? So what happens is when this object gets up to its maximum height, the velocity is going to be 0. So here the velocity final equals 0. Therefore, there is no kinetic energy and there is going to be some potential energy because now we're at some height above the ground here. So let's go ahead and expand out our terms. What I've got here is \( \frac{1}{2} m v_{\text{initial}}^2 = mg y_{\text{max}} \). So I'm going to call this \( y_{\text{max}} \) here and I'm going to go ahead and solve for this. Well, one of the things we'll notice here is that the mass is going to cancel. Usually, that happens with these kinds of problems. And now we're just going to go ahead and figure out \( y_{\text{max}} \). So \( y_{\text{max}} \) or \( y_{\text{final}} \) is going to be \( \frac{1}{2} v_0^2 \div g \). So if you go ahead and plug in our numbers here, you're going to have \( \frac{1}{2} \times 40^2 \div 9.8 \), and you're going to get, 81.6 meters. And that's the answer. So it goes 81.6 meters high, that's actually really high, it's like 250 feet or something like that. So if you could actually throw this thing up with, with that speed, and of course, there was no air resistance, that's how far it would go. Alright. So that's it for this one, guys. Let me know if you have any questions.
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10. Conservation of Energy
Intro to Conservation of Energy
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