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Ch 02: Kinematics in One Dimension

Chapter 2, Problem 2

A rock is tossed straight up from ground level with a speed of 20 m/s. When it returns, it falls into a hole 10 m deep. (a) What is the rock's speed as it hits the bottom of the hole?

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Hey everyone welcome back in this problem. A coconut is thrown vertically upward with a speed of 15 m per second from the top of a coconut palm 3.5 m above the ground and were asked to determine the velocity of the coconut when it strikes the ground. Alright, so let's just draw a little picture. We have our palm tree here, you can imagine that this is a palm tree, my terrible drawing And the palm tree is going to be 3. m tall. The coconut is going to be thrown from the top of the palm tree straight upwards, it's gonna travel up and then it's going to travel back down to the ground. And we want to know what the speed here is when it hits the ground. So we want to know this speed when it hits the ground. Now when we have a problem like this, we need to break this up into two stages. In the first stage I'm going to draw in Red is when the coconut is traveling upwards to the maximum height. The second stage I'm gonna draw in. Blue is gonna be when it reaches the maximum height and then comes back down to hit the ground. Okay, so there's two kind of periods that we need to look at. So let's look at the first one, the upward period. And I'm gonna write this in red to match the red on our diagram now in the upward period, we know that the initial speed, the knot up. Okay, we're gonna use you sub scripts to indicate the upward phase, We know that this is 15 m/s. We're told in the problem of that, is that initial upward speed. We know that the final upward speed is going to be 0m/s. Why is that? Well, at the very maximum height when the coconut goes from going upwards to going downwards, it's going to momentarily come to rest. Okay? So it's going to have a speed of zero right at that maximum height before it starts coming back down. So when we're looking just at the upward period, It's going to have a speed of zero m/s. The acceleration is going to be negative 9.8 m/s squared. Okay, the acceleration due to gravity and that's assuming that we're taking upward as positive. We don't know how long this phase lasts and we don't know how high the coconut. What distance the coconut reaches is delta. Y. You all right. Now, if we look at the downward period and that's where the information that we want to find is. So the initial speed in the downward well, this is going to be equal to 0m/s. Just like the final speed in the upward period, The coconuts gonna momentarily come to rest when it's at its maximum height before it starts to come back down. And so the initial speed in the downward period is going to be 0m/s. The final speed in the downward direction is what we're trying to find. That's what the question asked us to find. The acceleration is going to be the same negative 9.8 m/s square. The acceleration due to gravity and we don't know the distance or the time the coconut will travel. So if we can figure out how high the coconut goes in the upward period we will know how far it goes in the downward period. And that will allow us to calculate V. F. D. Okay right now we only have two knowns in our downward period. So we can't calculate V. F. D. So if we can figure out how high the coconut goes, we will know how far it's going to fall and then we can calculate the F. D. Alright, great. So we're gonna start with the upward period and we're gonna choose an equation that doesn't involve t Okay because we would like we have information about three variables and we want to find information about delta. Y. You okay And so that equation is going to be V. F. You square is equal to v not U squared plus two A delta Y. You okay subbing in the values we know we're gonna get zero on the left hand side is equal to 15 m per second. All squared plus two times negative 9.8 m per second squared times delta Y. You and this is going to give us a delta Y. You times. Okay if we move this to the left hand side, 19.6 m per second squared is equal to 225 m squared per second squared. And we can solve for Delta Y. You by dividing The 19.6 and we get Delta Y. U. is equal to 11.48 m. Alright, great. We know how far the coconut travels up. Now. We can use that to figure out the speed coming down. However we have to be careful here. Remember that the coconut started on top of the palm tree? Okay, so the distance has to travel down is actually more than the distance of coconut traveled up. If we look at our diagram, The Palm tree was 3.5 m tall. And we just found that the coconut traveled 11.48 m. Okay, so that distance is going to be here. This is Delta Y. You when we talk about Delta Y. D. That's going to be the entire distance to the ground. And so it's going to be the sum of Delta Y. U. and the 3.5 m. The height of that coconut tree or the height of the palm tree. Okay, so Delta Y. D. Is gonna be delta Y. You plus 3.5 m which is equal to 11.48 m Plus 3.5 m. And so it's going to be 14.48 m. So now we have information about three different things. V. Not A and delta Y. D. And we need to find V. F. D. All right. And be careful. I've said that this is the sum of these. That's actually negative delta Y. D. Okay, Delta Y. D. Is going down when the coconut is falling down. That's in the negative direction. And so the change in height is going to be negative. Um And so negative delta Y. De is 14.48. Alright, so now we're gonna use our um equation without T. That's the same equation. We use the first time. Okay? And we're gonna use it with these values. So we get V. F. D squared is equal to B not D squared plus two A. D. Delta Y. D. We're trying to find the F. D. Squared be not D. Is just zero. Okay, so this term is going to go away. We had two times negative 9.8 m per second squared times negative 14.48 because we're traveling downwards in the negative Y direction. So if we work out the left hand side, we get that V. F. D squared is equal to we have a negative times a negative so we get positive 293. m squared per second squared. And if you missed the negative on the 14.48 m if you forgot to add that in when you get to this stage, you would end up with a negative value here, you wouldn't be able to take the square root. Okay? So there would be some sort of warning, something's gone wrong and you'd have to go back and look at your work and try to figure out where where it went wrong. Okay. All right. If we take the square root we get that V. F. D. Is equal to positive or negative 17.135 m per second. Okay, When we take the square root we get the positive or negative value and we need to interpret which one is correct. Okay, well the coconuts coming down so the speed or the velocity is going to be pointing downwards. We've chosen upwards of our positive direction and so the speed or the velocity is going to be negative. So we have negative 17. m per second. Okay, that is the velocity of the coconut when it strikes the ground. If we go back up to our answer choices, we see that that corresponds with answer choice. A thanks everyone for watching. I hope this video helped you in the next one
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