(II) A rocket rises vertically, from rest, with an acceleratio...

Question

(II) A rocket rises vertically, from rest, with an acceleration of 3.2 m/s² until it runs out of fuel at an altitude of 725 m. After this point, its acceleration is that of gravity, downward.

(c) What maximum altitude does the rocket reach?

Accepted Answer

Hello, fellow physicists today we solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A stone is thrown vertically upwards which travels up to a height of 20.0 m with an acceleration of 1.2 m per second squared. And thereafter, its acceleration is equal to the acceleration due to gravity, determine the maximum height attained by the stone. So that's our end goals. We're trying to figure out what the maximum height that the stone will achieve when it's thrown vertically upwards into the air. Awesome. We're also given some multiple choice answers. They're all in the same units of meters. So let's read them off to see what our final answer might be. A is 24.9 B is 20.4 C is 2.44 and D is 22.4. So first off, let us consider the upwards motion to be positive and that the initial this stint which we're gonna denote as Y subscript I is equal to 0 m to be the ground level. So the initial distance is at the ground level and it's at 0 m. Now, we must recall and use the uniformly accelerated motion equations to solve for what the stone's velocity is when it reaches a height of 20.0 m. And let's for simplicity's sake, let's call this equation one. So the final velocity squared is equal to the initial velocity squared plus two multiplied by the acceleration multiplied by the change in distance. So at this stage, let us write down all of our known and unknown variables. So let's make a little note here that the initial velocity is equal to 0 m per second. And we do not know what our final velocity is. We know the acceleration is given to us as 1.2 m per second squared. And we know that the change in distance which is delta Y is equal to the final distance minus the initial distance, which in this case, it's equal to 20.0 m minus 0 m. So that just means that the change in distance is equal to 20.0 m. Awesome. So let's put a little box here. So it doesn't get jumbled up with all of our other notes that we're gonna be writing down for solving this problem. So now at this stage, we can plug in all of our known variables and solve. So since we do not know what the final velocity is, we're gonna be trying to isolate and solve for the final velocity. So let's do that. So the final velocity squared is equal to 0 m per second squared plus two, multiplied by 1.2 m per second squared, multiplied by 20.0 m for the change in distance. So that means that the final velocity squared is equal to 48 m squared per second squared. So to get rid of the square, so the velocity squared, we need to square root both sides. So when we do that, we will get that the final velocity is equal to 6.928 m per second. And this is because we square root both sides of the equation. Fantastic. So that means our final velocity has to be 6.928 m per second. At this stage, we can now begin to solve for our final answer for the maximum height of the stone. So this is what the. So basically, we're trying to find what the maximum height that can be attained by the zone, the stone when it's thrown straight up into the air. So note that the final velocity in this case will be zero since the stone's velocity will be 0 m per second for a split second before it falls back to the ground when it is thrown up in the air. So when it's thrown up into the air, right before the stone starts to fall back down to earth or back to the ground for a split second, it will be 0 m per second. So that's where that value comes from. So let's do like we did before. Let's draw another little box here off to the side and write down our known and unknown variables here. So we know that the initial velocity in this case, which will use the value we just found a second ago, which was 6.928 m for a second. So this is when the stone is thrown up to 20 0.0 m. And then as we stated before, our final velocity in this case will be 0 m per second. And the acceleration is gonna be equal to the acceleration due to gravity. So it's gonna be equal to negative 9.81 meters per second squared because once you throw the stone up and it starts to fall back down earth, its gravity is gonna be trying to pull it right back down to the ground. So it's gonna be accelerating at the rate of gravity. Awesome. And then we do not know we're trying to determine the maximum height attained by the stone. So the change in distance is equal to the final distance minus 20.0 m. So in this case, we are gonna be isolating and trying to solve for the final distance. So why final by itself? And this will make more sense in a in a minute. So using equation one again. So plugging in our known variables for this case back into equation one, we know that the final velocity is 0 m per second squared is equal to our initial velocity which is 6.928 meters per second square plus the Excel or I should say, don't forget, or two plus two multiplied by the acceleration due to gravity which is negative 9.81 m per second squared for acceleration multiplied by Y minus 20.0 m. Awesome. So like I said before, we need to isolate and solve for Y so with that in mind, you'd start doing that and say that Y minus 20.0 m is equal to 6.928 m per second squared divided by two multiplied by 9.81 m per second squared. So when we divide 6.928 squared divided by two multiplied by 9.81 we will get that Y minus two. So 20.0 m is equal to 2.4463 m. So no to solve for Y we have to add 20 m to the other side. So Y is equal to 2.4463 m plus 20.0 m which is equal to 22.4463 meters which is approximately equal to 22.4 m. And that is our final answer. Hooray, we did it. So looking at our multiple choice answers. That means the correct answer has to be the letter D 22.4 m. Thank you so much for watching. Hopefully that helped and I can't wait to see the next video. Bye.

Struggling with Physics?

(II) A rocket rises vertically, from rest, with an acceleration of 3.2 m/s² until it runs out of fuel at an altitude of 725 m. After this point, its acceleration is that of gravity, downward.
(c) What maximum altitude does the rocket reach?

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Struggling with Physics?

(II) A rocket rises vertically, from rest, with an acceleration of 3.2 m/s² until it runs out of fuel at an altitude of 725 m. After this point, its acceleration is that of gravity, downward.(c) What maximum altitude does the rocket reach?

Watch next

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(II) A rocket rises vertically, from rest, with an acceleration of 3.2 m/s² until it runs out of fuel at an altitude of 725 m. After this point, its acceleration is that of gravity, downward.
(c) What maximum altitude does the rocket reach?