A lunar lander is making its descent to Moon Base I (Fig. E2.40)....

Question

A lunar lander is making its descent to Moon Base I (Fig. E2.40). The lander descends slowly under the retro-thrust of its descent engine. The engine is cut off when the lander is 5.0 m above the surface and has a downward speed of 0.8 m/s.With the engine off, the lander is in free fall. What is the speed of the lander just before it touches the surface? The acceleration due to gravity on the moon is 1.6 m/s

Accepted Answer

Hey everyone welcome back in this problem. We have a moon lander case. So this is a spacecraft that's designed to land on the surface of the moon when it is 15 m above the surface, the engine is going to be cut off. Okay. And the spacecraft is going to have a speed of one m per second in the downward direction. When the engine cuts off, the lander is going to be free falling under the gravity of the moon. And we are asked to find the landing speed of the lander just before it reaches the surface. Okay? And we're told to consider the acceleration due to gravity on the surface of the moon as 1.6 m/s squared. Alright, so let's take down to be the positive Y direction in case we have our little diagram here and let's think about our um equations. Okay, we have free fall, we have uniform acceleration. Um So let's use our um equations. So let's figure out what we have, what we know, we know that the initial speed is going to be one m per second. We were told that in the problem. Now this is positive because we've taken down to be the positive Y direction and we're told that we're the speed is in the downward direction. Okay, so this is gonna be positive V. F. This final speed, what we don't know this, but this is what we want to find. Okay, we want to find the speed just before it reaches the surface. So V. F. That's what we want to find. Okay, what about acceleration? Well, the acceleration we're told is 1.6 m/s squared on mars or sorry, on the moon. We're talking about acceleration due to gravity And we know that this is also in the downward direction which we've chosen as positive. So this is gonna be positive. 1.6 m/s squared. Okay, Delta Y. The change in height or the change in distance? In the Y direction? This is 15 m. Okay, we're starting 15 m above the surface and we want to know what's happening when we get to the surface. Okay. So our change in Y is gonna be 15 m were not told anything about T. But that's okay because we already have three of the other variables. So we can go ahead and choose the equation without T. Okay. And that equation is going to be the following. The f squared is equal to v naught squared plus two A delta Y. So filling in the information. We know we have V F squared on the left is equal to v naught squared. Well, that's one meter per second elsewhere. Plus two times a 1.6 m/s squared Times Delta Y. Which is 15 m. Alright, so we have V F squared. Okay, one m per second squared. Well this is just gonna give us one m square per second square two times 1.6 times 15. This is going to be 48 in our unit here meter per second squared times meter. So we also have meters squared per second squared, so we have the same units. We can go ahead and add, we get V f squared is equal to 49 m squared per second squared. Okay? And taking the square root, we're going to get V. F is equal to positive or negative seven. And the units when we take the square is meters per second, which is what we want for velocity. So that's great. Alright, now we have the positive and the negative route and we need to kind of figure out based on the question which route we should be taking. Okay, well, we know that our spacecraft is going downwards. The acceleration is going downwards, its initial speed is going downwards, so its final speed is also going to be moving downwards. So that's the positive direction. So we're going to take the positive route. So VF is equal to 7m/s. Okay, That is going to correspond with answer D. That's it for this one. Thanks everyone for watching. See you in the next video

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