You throw a rock off the top of a tall building at an upward angle of 15°. At t=3 s, the rock's horizontal displacement from you is 52m. How high does the rock get above the top of the building?

Hey, guys. Let's take a look at this problem and work this one out together. So we're throwing a rock off of the tall building and some angle, and we're told, that something about the time the displacement, and we're figuring out how high the rock goes above the top of the building. So let's go ahead and draw a quick little sketch. So we know we're drop we're throwing something off of the roof of a building. So we're gonna throw something off right here at some angle like this, and which means it's gonna take an upward path like this. It's gonna come back down and then it's gonna keep going a little bit further. So let's go ahead and draw the path in the x and y, our path in the x look like this, path in the y will look like this, and then our points of interest. So we know our initial, we know that the maximum height is a point of interest, then the height, that's point C. And then we have another one that's point D, the part where it hits the ground. So here is our path in the x axis. So from b to c to d, and then it goes up to b, down to c again, and then down all the way to D. So that's the first step. So the second step is now we just have to figure out the target variable. What are we looking for? We're looking for how high the rock gets above the top of the building, which is the location which it was thrown. So basically, we're looking for the distance, the vertical distance that it goes between a and b. So this is gonna be delta y and we're looking at the interval from a to b. Right? So we always wanna include point b here. So that means we're looking for Delta Y from a to b. Remember this b point is special. It's kind of important because we know that VB in the y axis, oh, oh, is gonna be 0. So here at the top of the peak, there is no y velocity and that's why we like using that point. So I'm gonna use my 5 y variables. I've got ay as negative 9.8. I've got v naught y which is v a y, which I don't know. I've got v y, the final velocity vby, I know that's 0. And Then I've got delta y from a to b which is what I'm looking for. And then I've got the time from a to b. Okay. And I don't know what that is either. So, it looks like I have 2 out of 5 variables. I mean, we need to figure out either v a y or t a b. Now when I'm stuck, I usually just look at this time variable, and then I just go to the y the x axis. So if I'm ever stuck, you can figure out the time variable by going to the x axis. So let's see if we can do that here. So I'm gonna go ahead and go to the x axis. So my x axis and I'm looking for time from a to b, and that's my target variable. And so I'm gonna use my only equation delta x from a to b equals vx times ta to b. And let's see. Do we know delta xab? No, we don't. Do we know the initial velocity in the x axis? No, we don't. And then we're trying to look for time here. So it turns out that this equation is actually terrible to use because there's 3 unknowns here. We don't know any one of those values. So it turns out that going to the x axis is not really going to be our best bet. So if this is the case, if we can't solve the time using the x axis and it looks like we're stuck again. We're trying to figure out this delta y here and so our only option is to figure out this y velocity over here. What is the v a y? So what is that in our diagram? Well, let's take a look here. The only thing we're told about the initial velocity is that the upward angle is 15 degrees. So what this means here is that if we wanted to figure out v naught y, which is v a y, which is just v naught times the sine of theta, I have the theta. All I need to do is just figure out this initial velocity in the y axis. So how do I do that? What's my other components? I have v naught x which is just v ax, which is vx, and that's gonna be v naught times the cosine of theta. Okay. So if I again, again, you know, if I can figure out this v naught here, then I can bring it back into this equation and figure out v a y. So what are we told in the problem here? The only thing we know about the time and the initial velocity, well, the only thing we know about the initial velocity is the upward angle is 15 degrees. The only thing we know about time is that at t equals 3, the rocks horizontal displacement from you is 52 meters. So what does that mean? Well, here's what I'm gonna do. I'm gonna just draw an imaginary point. I don't know exactly where it is in the trajectory, and I'm just gonna call this point point p. And so what I'm gonna do is in the x axis, I'm gonna draw this point p and all I know about this point is that the time that it takes to get to point p is 3 seconds, and I know the delta x, so that means I'm gonna call this delta x from a to p, is 52 meters. Now why is this useful? Because remember that vx, the velocity in the x axis, is never going to change. And so one of the other equations that we can write for vx is that it's delta x over delta t. At any point in time, if you know delta x and t, then you can figure out the x velocity. So we know the delta x from a to p, and we all know the time from a to p. So this is just gonna be 52 over 3, and so we get 17 point 3 meters per second. And remember, the 17.3 meters per second will never change throughout the problem. It's always gonna be that number. So why is this useful? Because we know that we're if we're trying to find v naught here and we have the theta angle and we have 17.3, and what we can actually do here is we can say that v naught is equal to 17.3 divided by the cosine of 15 degrees, the angle. And if you do that, we're gonna get a launch velocity of 17.9 meters per second. And so now we can plug it back into this equation here and say that v naught y is gonna be 17 point 9 times the sine of 15 degrees, and that's gonna be 4.63 meters per seconds. So this is all really just to get this v a naught a v a y term, which is gonna be over here. So this is gonna be 4.63. So we jumped around these equations to find v a naught. And now now we have 3 out of 5 variables, we're just gonna stick to the normal steps. Right? So we have this time variable is gonna be my ignored and so we're just gonna pick an equation that it uses, that ignores time. So if I wanted to find delta y from a to b, then what I'm gonna use is I'm gonna use equation number 2, which says that vby squared is equal to vay squared plus 2 times ay times, deltay from a to b. So this is my target variable. And I know that vby is equal to 0, so this is gonna be 4.63 squared plus 2 times negative 9.8 times delta y from a to b. So if you go ahead and solve for this, what you're gonna get is that delta y from a to b is equal to 4.63 squared divided by 2 times 9.8. And if you go ahead and work that out, you're gonna get 1.1 meters. So that is the answer. That is the vertical displacement that the rock takes to go from A up to B. So we know that this distance here is 1 point 1 meters and that's delta y. So that answer choice is Andrew Joyce a. So hopefully this makes sense guys, and let's just go ahead and get to the next one. So let's sit for this.

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You throw a rock off the top of a tall building at an upward angle of 15°. At t=3 s, the rock's horizontal displacement from you is 52m. How high does the rock get above the top of the building?

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