A ball is thrown from the top of a 50-m-tall building with a speed of 40 m/s at an angle of 37° above the horizontal. How far horizontally does the ball travel before hitting the ground?

Hey, guys. Let's check out this problem here. We've got a ball that's thrown from the top of a rooftop, on top of the building, and we're told the initial speed and the angle, and we're gonna figure out how far horizontally the ball travels before it hits the ground. So let's just go ahead and draw a quick little sketch. I've got this building top here and then the ground is gonna be over here. The ball is gonna be thrown with some initial velocity. And because it's thrown upwards, it's gonna take this path like this, and then eventually, it's gonna hit the ground at some later time. So let's go ahead and stick to the steps. We're gonna draw the path in the x and y. So our path in the x is to look like this. Our path in the y is gonna go up and then come back down again. So our points of interest, we would a, and the maximum height is point b. Then we've got the initial height, place where basically it just becomes symmetrical, which is c, and then finally down when it hits the ground at point d. So I'm just gonna go ahead and fill out all the, all the paths in my x and y axis, a, b, c, and d. Okay. So we've got that. And now we're just gonna figure out the target variable. What are we looking for here? We're trying to figure out how far horizontally the ball travels before it hits the ground again. So that's, because we're looking for a distance in the horizontal axis, we're looking for a displacement delta x. So now which interval are we gonna use? Well, remember, we're always gonna try as best as we can, just use the interval, just a single interval from a all the way to d. And if we can't figure that out, then we'll probably have to come back here and then figure out another interval. Right? So you always should always try the interval, a single interval from a to d and see if that works. So that's the interval we're gonna use. And then what's the UAM equation? Well, we're looking for delta x. So we're just gonna use the only delta x equation. So we're looking for delta x from a to d. We need vx times t from a to d. Alright. So let's take a look here. So if I wanna figure out delta x, then I need vx and then I need t. So what's what's vx? Well, if I had the initial velocity, if I had this v not here and I'm told it's 40 and I have the angle, which is 37, I've got a magnitude and direction and I can figure out v not x, which is just v ax, which is just vx throughout the whole thing. And that's just gonna be 40 times the cosine of 37, and you're gonna get 32. And then while we're at it, we'll get the initial velocity in the y axis, v a y, That's just 40 times the sine of 37, and that's gonna be 24. So we've got our x and y components of the initial velocity. So this is 40 over here. Cool. So we know what this vx component is. What about the time? I'm gonna need t a d if I wanna figure out the horizontal displacement. But I'm not told anything about time in the problem, so it looks like I'm kinda stuck here. Remember when I get stuck, I'm always just gonna go to the y axis and then figure out the time in the y axis. So I'm gonna try to figure out TAD. So I'm gonna need my 3 out of 5 variables. I'm gonna need a y, which I know is negative 9.8, so no matter what interval I'm in. And my my initial velocity in the y axis, which is v a y, which is just calculated over here. That's just 24. The final velocity, remember my interval is from a to d, which means that this gonna be the final velocity at point d. I don't know what that is. Delta y from a to d and t from a to d. Remember that I came over here to the y axis because I need this time here. So I'm gonna need one of these other two variables either vdy or delta y from a to d. So if we take a look at our problem here, we don't know anything about the final velocity. But we do know, however, that the building is 50 meters tall. So what happens is if we're going up to point b down and then all the way down to the ground again, effectively, from the interval from a to d, you've just gone downwards like this. And so this is just gonna be your delta y from a to d and that's gonna be negative 50 because you're going downwards. Right? So we have negative 50 over here. Okay. So we don't know what this is, but we have 3 out of 5 variables, so that's okay. And we're just gonna go ahead and pick the equation that does not include, the final velocity. So that is the equation number 3, and that says that delta y from a to d is equal to vaytadplusonehalfaytadsquared. So now we have everything, let's just go ahead and plug in all of our numbers here. Delta y from ad is negative 50, and this is gonna be 24 times tadplus, and then one half of a y is just gonna be negative 4.9 Tad squared. So again, I still only have one unknown here. I've got this tad, but unfortunately, what happens is I have a power of tad, you know, I have a term to the first power and then term to the second power. So I've actually ended up with a quadratic term over here. So I'm gonna have to rearrange it. And so I'm gonna rearrange this so so that it looks like 4 negative 4.9 t a d squared plus 24 t a d plus 50 equals 0. That's what happens when you bring 50 to the other side, because then it looks like a t squared plus bt plus c equals 0. We know what to do here. We're just gonna use the quadratic formula. So unfortunately, we're gonna have to use the quadratic formula to figure out time. What are our roots for time? What are the two solutions? Remember, that's just gonna be t. It's gonna equal to, it's gonna give you 2 solutions, negative b plus or minus the b squared minus 4ac. They're all divided by 2 a. So what happens is, we're gonna get negative 24 plus or minus the square roots of 24 squared minus 4 times negative 4.9 times 50. So you go ahead and do that and you're going to divide this by 2 times negative 4.9, and you're gonna get the 2 solutions out of that. So t can either be, let's see. I get negative 1.58, and then I get positive 6.47. Now it doesn't make sense to have a negative time. So this is actually gonna be, our correct time. It's 6.47. So now I take this number and I plug it all the way back into my initial expression in the x axis, and then I finally get that my delta x from a to d is equal to vx, which is 32. Remember, that's what I figured out over here. My x component of my initial velocity, which never changes, and then times 6.47. So if you go ahead and work this out in your calculators, you're gonna get a horizontal displacement of 207 meters. So that is the answer. We scroll up to our answer choices and that's gonna be answer choice c So that's it for this one, guys Let me know if you have any questions, and I'll see you in the next

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A ball is thrown from the top of a 50-m-tall building with a speed of 40 m/s at an angle of 37° above the horizontal. How far horizontally does the ball travel before hitting the ground?

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