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Ch 02: Kinematics in One Dimension

Chapter 2, Problem 2

As a science project, you drop a watermelon off the top of the Empire State Building, 320 m above the sidewalk. It so happens that Superman flies by at the instant you release the watermelon. Superman is headed straight down with a speed of 35 m/s. How fast is the watermelon going when it passes Superman?

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Hey everyone in this problem, a tennis ball is released without an initial speed from the top of a skyscraper. The height of the skyscraper is 830 m at the instant the ball was released. A drone located in the same position as the ball was moving vertically downward With a speed of 23 m/s and were asked to determine the speed of the ball when it catches the drone. Alright, we're asked for the speed of the ball when it catches a drone. So let's start by listing out the information we know about the ball. Yes, we have a ball. What do we know we know that its initial speed or initial velocity It's going to be equal to zero. Why is it zero over told that it's released without an initial speed? Okay. So the initial speed is 0m/s. The final speed of the ball. This is what we're trying to find. Okay. The final speed of the ball is what we're gonna say is the speed of the ball when it catches the drone and this is gonna be velocity because we have that vector symbol velocity of ball when it catches the drum. Alright. So that's what we're looking for. What else do we know? Well, we have acceleration due to gravity. Okay. We're gonna take the upward direction as positive. And so when this ball is falling from that skyscraper, the acceleration is going to be negative and we're gonna say subscript B for the ball negative 9.8 m per second squared. Now, we don't know how far it's traveled and we don't know the time. Now, let's think about this. We only have two knowns. We have something we want to find and we have two other unknowns. So we can't use our um equations like this. We don't have enough information. What we do know is we want the speed of the ball when it catches the drone. Okay. So when the ball catches the drone, delta Y for the drone and the ball will be the same, they start in the same place. So when it catches it, they're going to be in the same place. And so the change in height is going to be the same, the time is also going to be the same. Okay. When it catches the drone, we're gonna be at the same time. So let's let delta Y C and T C represent the values when the ball catches the drone. Okay. So the C kind of represents catch, okay. So we have delta Y C and T C here because we want to know the speed when the ball catches the drill. All right. What about the drone? What's right? Information about the drone on the right hand side here. Now, we're told that drone is moving vertically downward with a speed. Okay. So it was a constant speed. If it has a constant speed, then we only need to worry about its speed. It's change in height and the time. Okay. We don't have to worry about the acceleration, it's constant speed. So the acceleration of zero and we don't have to worry about an initial or final velocity because the speed is constant has the same speed throughout. Alright, so the velocity of the drone, hey, the drone is moving downwards again, we've chosen up to be our positive direction. And so this is going to be negative 23 m/s. We don't know how far it travels or how like how far down it goes. So delta Y C is unknown and TC is also like not. Alright. So we have information about the ball, we have something we want to find and we have two other unknowns. And then for the drone, we have one known in two unknowns. So we aren't able to solve for all of these variables. Okay. We have three variables we don't know and we have only two equations we can write off the back. So let's go ahead and start with delta Y C and T C because we have those in both equations. So we have two variables and we can write an equation for each of the ball and drone two equations, two unknowns will be able to solve. All right. So if we're looking at the ball and we're looking for an equation that does not include that final velocity. Okay, we're gonna choose the following um equation. Delta Y C is equal to V, not B times T C plus one half A B times T C squared. Okay. So the change in height is equal to the initial velocity times time plus one half the acceleration times times square filling in what we know we have. Delta Y C is equal to well, V not B A zero. So this term goes to zero, we're left with one half times the acceleration of the ball negative 9.8 m per second squared times T C squared. And if we just simplify this one half times negative 9.8 m per second squared, we get that delta Y C is equal to negative 4.9 meters per second squared times T C squared. And we're going to label this as equation one because we're gonna need to come back to it. Now, let's do the same for the drone. We're dealing with the drone that has a constant speed. And so the equation we're gonna use is just a velocity is equal to distance over time. And so we have V D is equal to delta Y C over delta over T C. Okay. And if we want to write this as an equation for delta Y C, we can write that delta Y C is equal to negative 23 m per second times T C. And we're gonna call this one equation star. Alright. So now we have two equations with two unknowns. We have delta Y C and T C in both of these. So let's go ahead and use our substitution method to solve for one of these values. So we want to sub equation star into equation one. So equation star tells us delta Y C is equal to negative 23 m per second times T C. So in equation one, the left hand side, delta Y C is going to be replaced by that expression. So we're gonna have negative 23 m per second times T C is equal to negative 4.9 m per second squared times T C squared. And we can solve, let's move everything to the left hand side. We got 4.9 m per second squared times T C squared is equal to is not equal to minus 23 m/s times TC is equal to zero. Now, this is a quadratic, we don't have a constant term. So we can factor out A T C. So we have T C times 4. m per second squared times T -23 m. /s. All of this is equal to zero and this should be T C. Not just T okay. Now, we can have either of these factors equal to zero. So we have that T C is equal to zero or 4.9 m per second squared times T C minus 23 m per second is equal to zero. Okay. This tells us that T C is equal to m per second, divided by 4.9 m per second squared, which is going to be equal to approximately 4.6939 seconds. Now we found two time values. We have to think about which one we should be using. Well, the first time value is zero. That makes sense. We know at time zero rate is at balls drop that they're at the same position, they start at the same position. So T C equals zero represents that starting position. Okay? And this second time is going to represent the time where the ball catches the drone once it's been dropped. So now we have this time value. Let's go back up and look at what we're trying to find. Okay. Remember that we're trying to find this speed of the ball when it catches the drone. This VFB. Okay. Now we have a value for T C we have that this is 4.6939 seconds. Now, if we look at the information, we have three notes, the initial speed, the acceleration and the time. So we can solve for VFB. Now with, are you a M or Kinnah Matic equations? Okay. So we're gonna choose the equation without delta Y C because we didn't find delta Y C, we're gonna sub in our information. So we have the equation V F B K. The final speed or velocity is equal to V not be the initial velocity plus A B, the acceleration of the ball times T C that time that we were interested in when the ball catches a drum. Okay. On the right hand side, we know that the initial velocity is 0m/s. The acceleration is negative 9.8 m/s squared. And the time is 4. seconds. If we work this out on our calculators, we get That the velocity of the ball is negative 46 m/s. Okay. We were asked to find just the speed, Okay. So we want to find just the magnitude of that velocity. We don't care about the direction. The negative does indicate that the ball is going downwards because we've chosen up as our positive direction. So that makes sense. The ball is falling down. Our speed is going to be the absolute value of that velocity. The magnitude which is going to be 46 meters per second. Okay. And that is that speed that we were looking for. If we go back up to our answer choices. We see that we found that the speed of the ball when it catches the drone is going to be c 46 m per second. Thanks everyone for watching. I hope this video helped see you in the next one.
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