You launch a rocket with an initial speed of 5×1035\times10^3 m/s from Earth's surface. At what height above the Earth will it have ¼ of its initial launch speed? Assume the rocket's engines shut off after launch.

1.47×1061.47\$$times10^61.47×106 m$$

You launch a rocket with an initial speed of 5 × 10 3 5\times10^3 m/s from Earth's surface. At what height above the Earth will it have ¼ of its initial launch speed? Assume the rocket's engines shut off after launch.

Hey, guys. Hopefully, you're able to try this one on your own. So we've got a rocket that we're gonna be launching from a launch pad on the surface of the earth. So I've got this rocket that's gonna be right here. We've got initial velocity and that's gonna travel some height, h, and at that point, it's gonna have some final velocity, v final, that's gonna be a quarter of the initial velocity. And our job is to figure out exactly how high that is so that it gets to this point and it has that velocity. So let's go. We've got h equals our target variable. And every time for solving for h, we're always gonna solve it using little r equals big R plus h. But we've got 2 different points that we're talking about. First, it's on the surface and then we're at some higher distance here, and we have a larger velocity. We have large v. So that means we're gonna use energy conservation. Right? So we've got energy that we're working with. So let's set up our energy conservation equation. We got the initial kinetic and potential, any work done, and that's equal to final kinetic and potential. So let's see. Once this thing takes off and the engines shut off after launch, that means the force of gravity is working on this thing and that's conservative force. So that means that there's no non conservative works done. And then let's go ahead and substitute all of these expressions in. We've got one half, the little m times vi squared. And then how about the potential? Well, let's see. Even this thing even though this thing is on the the surface of the earth, there's still some center of mass distance, little r I, away from the rocket. So that means that which is the radius of the earth, by the way. So that means that there's still some gravitational potential. So I can actually go ahead and write this expression as g big m, the mass of the earth, little m over big r, which is the radius of the earth. And then we've got, finally, at this final position, we've got 1 half mvfinal squared minus gbigmlittlem over well, let's see. At this point, the center of mass distance is this entire piece right here, r f. And if I'm looking for h, that's my target variable, I can always relate that using r f. So that's actually gonna be my target variable. So this r little r f is really what I'm looking for. And I can if I can find out that, then I can plug it back into this expression and then figure out what that little h is. So that's the game plan. So let's go ahead and see if I can cancel out anything from my energy conservation equation. I've got little m's that pop up everywhere, which is good because I actually don't know what the mass of the rocket is. And then if I want this RF by itself, I'm gonna start moving everything over to one side. So I've got this v final, this one half v final squared term here. I'm gonna move over to the other side. So now, basically, I've got one half vi squared minus g bigm over r minus 1 half v final squared equals negativegm over little r f. All right? So this is, again, my target variable here. So if I was trying to solve for this little r f, I might be tempted to sort of like start manipulating all these variables and see if I can come up with an expression. The problem is I can't simplify this whole entire left term here. And if I try to flip this whole thing, it's gonna get really, really ugly. So instead, what I'm gonna do is I'm gonna see if I can actually simplify anything by just plugging in some numbers. So I have what initial velocity is. I know what those terms are, and then these are all just constants that I have here in the table. And then I've got the final velocity, which is just a quarter of the initial velocity. So what I'm going to do is instead of having to manipulate all the variables, I'm just going to plug in stuff and start getting numbers because I can simplify numbers. So I've got 1 half, then I've got 5 times 10 to the 3rd squared minus 6.67 times 10 to the minus 11, mass of the earth 5.97 times 10 to the 24th, Mass of the earth, 5.97 times 10 to the 24th, divided by the radius of the earth, 6.37 times 10 to the 6th. And then we've got minus 1 half, And then the final velocity is gonna be a quarter of the initial velocity. So 1 fourth times 10, 5 times 10 to the 3rd, square that. And then that's gonna equal negativegm over little r f. So now if you actually start plugging in all of these terms here, you're just gonna get numbers for those. This is gonna turn into 1.25 times 10 to the 7th. This term is gonna turn into you gotta remember the minus sign, 6.25 times 10 to the 7th. There's a minus sign. And then we've got this term becomes 7.81 times 10 to the 5th here. And that's gonna equal negative gm over r f. So if you actually just do all the subtraction for all these three terms, you're just gonna end up with one number, negative 5.08 times 10 to the 7th. That's equal to negative gm over r f. You see how much simpler that became. So now instead of having to do all of this work with all these variables, all we have is just one number, and then we can go ahead and solve for this little r f f by first canceling out the minus signs here and then going ahead and switching the positions of these 2. Right? So the r f goes up top and the 5.08 goes goes to the bottom. Right? So now we've got that the RF, the final center of mass distance here is just 6.67 times 10 to the minus 11. Then we've got 5.97 times 10 to the 24th. And then we've got 5.08 times 10 to the 7th on the bottom. So we're going to end up getting 7.84 times 10 to the 6th. But again, we're not done yet because remember we have to plug it back into our r +h equation. So for the final height, it's going to be, let's see, we've got r equals r+h. So rfinal equals r+hfinal. So that means hfinal is going to be, r minus r. So we've got 7.84 times 10 to the 6th minus the radius of the earth, 6.37 times 10 to the 6. So our final answer is 1.47 times 10 to the 6th and that's in meters. Let me know if you guys have any questions with that.

Difficoltà con Physics with Algebra?

You launch a rocket with an initial speed of 5×1035\(\times\)10^3 m/s from Earth's surface. At what height above the Earth will it have ¼ of its initial launch speed? Assume the rocket's engines shut off after launch.

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