A rock is thrown horizontally with a speed of 20 m/s from the edge of a high cliff. It lands 80 m from the cliff's base. How tall is the cliff?

Hey, guys. Hopefully, you got a chance to try this one out on your own. Let's check this one out together. So we've got a rock that's thrown horizontally off of a cliff, lands 80 meters from the base, and we're gonna calculate how tall the cliff is. Now before we start writing a bunch of equations and numbers, let's just draw a sketch of what's going on here. So we got a rock that's thrown perfectly horizontally off of a cliff, so like this. And so once it's given an initial velocity of this, then it's gonna take some parabolic path because it's under the influence of gravity. So now the points of interest. Remember the, first point of interest is always just to be your initial point, point a. And then anything else that might happen in between like a maximum height or anything like that. But what happens in this problem is that we throw the rock and then eventually it just hits, the ground 80 meters away from the rock's base. So that's basically the only points that we're gonna look at for this problem just from a to b. So now we're gonna draw the paths in x and y. In the x axis, it would look as if you just went straight across like this and at the y axis, it would look like you just dropped from a down to b. So these are our points over here of interest. That's the first step. So now we're gonna look at the target variable. What are we actually asking for in this problem? We're asking how tall is the cliff. So which variable is that going to be? Remember, we have to pick one of the target variables that's gonna be one of our equations. So which one is that? Well, how tall the cliff is is basically gonna represent this distance over here which is a which is a vertical distance or vertical displacement. So it's not gonna be delta x, it's not gonna be velocity or time, it's not gonna be any of these velocities, but it is going to be our delta y. Basically the target variable that we're looking for here is the vertical displacement from a to b of the rock. There we go. Okay. So that's gonna be our target variable. We're looking for delta y. So now that moves that that brings us to the 3rd step, which is we're gonna determine what interval we're gonna look at and then we can start figuring out which equation we're gonna use. So the interval we're look looking at is the one from a to b. We're gonna look for how tall or what's the vertical displacement between a, and the ending point which is b. So that means that we're using the interval a to b. Alright. So that means we're looking for delta y from a to b. And so now instead of starting in the x axis, like I would, if I were solving for time or something like that, I'm actually just gonna go to the y axis because I have a y axis variable. So what are my y variables? This is gonna be a y which I know is 9.8. My initial velocity which is just v a y which I remember in from horizontal launches is always gonna be 0 because if I have this velocity here, initial velocity is 20, then this is actually what the v x is and my v y knot component is just 0. All velocity is purely in the x axis. It's just only in the x axis. There's no component in the y axis. So what about the final velocity? That's gonna be the velocity at point b. I don't know what that is, but I do know that at the bottom here, it's gonna have some some vertical components. And then my last two variables are delta y from a to b which is what I'm looking for. And then I've got t from a to b. Okay. So I've got these 2 out of 5 variables. I need one more. So let's take a look. Do I know the final velocity in the y axis? Do I know v b y? I don't. And the only other variable that I could look for is time. So it looks like I'm a little bit stuck here because I only have 2 out of 5 variables, but I need 3 out of 5 in order to pick one of these equations. So again, remember, what happens is whenever you get stuck, you're just gonna go to the to the other axis. Here, we started off at the y axis. So now what I have to do is I have to go over to the x axis and I wanna solve for one of these variables. Which one makes sense? Well, I can't solve for the velocity in the y axis by going to the x axis. But I can solve for the x axis Or sorry, I can't solve for time in the x axis because it's the only one of the variables that I can solve in either the x or the y. So I get stuck here and I'm gonna have to go to the x axis because I'm looking for tab. Alright. If I can figure that out, then I can go back into this equation here. So the only equation I can use is delta x from a to b is vx times t a to b. So if I can figure this out, I'm good to go. So what is the delta from a to b? What's the horizontal displacement? Well, the only thing I'm told in the problem about horizontal displacements or any kind of horizontal distance is that it lands 80 meters from the cliff's base. So that means that if this is the cliff's base over here, then this delta x from a to b is just equal to 80. So that means I know that this is 80 and what about the horizontal velocity in the x axis? Well, remember that if you have this horizontal if you this initial velocity here, my v naught x, which is just v x, is just v initial, which is the 20 and then my v naught y is just equal to 0. So I have basically the v x, which is gonna be the same throughout the entire motion, it's never gonna change and it's gonna be 20. So that means that my t a to b is just 4 seconds. And so now I have what t a to b is. Now I can bring it back into this equation here. I know this is 4. And so, therefore, now I have 3 of my 5 variables. So now I can pick an equation that does not have my ignored variable, which is the final velocity. So if we take a look at our y equations, the the equation that doesn't have final velocity is the 3rd equation. So I'm gonna use equation number 3, which is delta y, and I'm gonna plug in all my subscripts. Right? So delta y from a to b is gonna be the initial velocity which is v a y times t a to b, plus 1 half a y times t from a to b squared. Alright. So that's the equation. One of the good things about horizontal launches is that we know this term is always going to go away because v naught v a y is equal to 0. So that term always goes away which is good for us. So delta y from a to b is just 1 half times negative 9.8 times t which is 4 squared. If you go ahead and work this out and plug this into your calculator, you're gonna get negative 78.4 meters. So this is, the answer to the vertical displacement. However, so notice if you look at your answer choices, we're looking for the height of the cliff or how tall the cliff is. So a lot of times what you'll see is that these heights are just the positive numbers or the absolute values of the numbers that you'll get. So instead, we'll get for the height of the cliff is 78.4 meters positive, because it doesn't make sense to say that it's negative 78.4 meters. Alright? So that's our answer choice over here. That is answer choice a and let me know if you guys have any questions.

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A rock is thrown horizontally with a speed of 20 m/s from the edge of a high cliff. It lands 80 m from the cliff's base. How tall is the cliff?

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