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Launching Up To A Height

Patrick Ford
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Hey guys, So let's go and check out this problem here, you're gonna launch a four kg object directly up from the ground so that the ground like this, you're gonna take this four kg ball and you're gonna throw it up with some initial speed. V not equals 40. And we want to figure out basically what happens when the ball finally comes to a stop here at some maximum heights. So here, if we take the ground level to be zero, what we want to calculate is what is this? Y max value here. All right. 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 gonna write our energy conservation equation. So we're gonna write K initial plus you initial equals K final plus you final. Now we're going to eliminate and expand out each one of our terms here. All right. So, we've got some initial kinetic energy. That's the initial speed of 40 m per second. So, we've got that. But here, when we're at the ground level, if we take y equals zero to be the grounds, right? We're starting from that means that our gravitational potential energy is zero here. So, therefore there is no gravitational potential. All right. So, there's nothing there. What about K final? So, what happens is when this object gets up to its maximum height, the velocity is going to be zero. So here the velocity final equals zero. 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 our terms. What I've got here is one half M V initial squared equals And then the gravitational potential energy is going to be M G H. Or rather MG Y max. So, I'm gonna call this Y max here. I'm gonna go ahead and solve for this. Well, one of things we notice here is that the mass is going to cancel. Usually that happens to these kinds of problems. And now we're just gonna go ahead and figure out why final? So sorry, why max? So why max or why final is going to be one half of V not squared divided by G. So you go ahead and plug in our numbers here are going to have one half of 40, squared divided by 9.8. And you're gonna get um 81.6 m. And that's the answer. So it goes 81.6 m high. That's actually really high. It's like 250 ft or something like that. So, if you can actually could throw this thing up with with that that speed. And of course there was no air resistance. That's how far it would go. All right, so that's if this one guys let me know if you have any questions.