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Before working this problem, review Conceptual Example $15 .$ A pellet gun is fired straight downward from the edge of a cliff that is 15 $\mathrm{m}$ above the ground. The pellet strikes the ground with a speed of 27 $\mathrm{m} / \mathrm{s}$ . How far above the cliff edge would the pellet have gone had the gun been fired straight upward?

22$m$

Physics 101 Mechanics

Chapter 2

Kinematics in One Dimension

Motion Along a Straight Line

Rutgers, The State University of New Jersey

Simon Fraser University

Hope College

University of Winnipeg

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okay, in this problem, we have somebody who's firing a pellet guns straight down off a 15 meter tall cliff. Um and we know the final velocity the pelican has before it hits. The ground is 27 meters per second. So is going down where? The 27 meters per second that height is 15 meters and the acceleration here is 9.8 meters per second squared. Thanks to gravity. Um, really quickly, because this problem has both ups and downs. I'm gonna define my axes because I'm starting off firing something straight down, actually going to define downwards as a positive value, which means anything that's going up will be a negative value. Um, this is useful because we know the pellet is actually ending below our starting point. So if we do find it the other way, kind of the more traditional way, all of our numbers here would have to be negative. And that would be a big headache when we get to the actual math side of things. Rather, I have a final velocity of 27 meters per second downwards on acceleration of 9.8 meters per second, squared downwards on the displacement is 15 meters downwards. Now, my actual goal here is to figure out, um, how high up the pellet would have gone if I had fired it straight up instead of straight down. So I need to do this in two steps. The information I'm giving here is enough to get me my initial velocity of the palate. That's what we'll use to figure out how high up the palate actually gets in our solution of the whole problem. Now, for the first part here, we know three numbers were looking for the initial velocity. We don't know the time, and it's not what we're looking for. So we're going to use the equation over here that does not have time inside of it, which is that v squared equals V, not squared, plus two a X. And my goal here is to figure out what that initial velocity is going to be. So I'm gonna plug in my numbers. We have 27 squared is equal to the initial velocity squared, um, plus two times 9.8 times 15 which gives us plus 294. Um, so when I collect my like terms here and kind of switch it so it looks a little nicer. We get V, not squared equals 435. And when I square root both sides, we have the initial speed of the palate as 20.8 meters per second. Now that again is the chunk we're going to be using to figure out how high up the pal it's going to get if we fire it directly upwards, um, and see how high it will reach. So let's add in another of those Kinnah Matic setups here my five variables that I'm looking at and this time we're using them to analyze the bullet's path as it goes upwards, not downwards so that 20.8 meters per second is now pointing up, which I said up is my negative value. So we have negative 20.8 meters per second. The acceleration is 9.8 meters per second squared downwards, so that stays the same. And because I'm looking at firing it straight up, my final velocity will reach zero meters per second. When my pellet gets to its maximum height, my goal here is to figure out what that maximum height is so a kin. I'm no using the equation that does not have a tea in it and is also not looking for time. So the squared equals V not squared. Plus two a axe. When I plug, my number's in here. We have zero is equal to that initial velocity squared, which gives us 432 0.64 plus 19.6 X. Um, then I'm going to subtract that over. So I have negative 19.6 x equals 432 0.64 and we divide by that negative 19.6 x to 19.6 to get a maximum height of negative 22.1 meters again to explain that negative, that's just our direction, saying this will go 22 22.1 meters upwards for its maximum height, and that is our final answer.

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