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Problem 31 Medium Difficulty

A quarterback throws a football with angle of elevation $ 40^\circ $ and speed 60 ft/s. Find the horizontal and vertical components of the velocity vector.

Answer

horizontal $\approx 45.96 \mathrm{ft} / \mathrm{s} \quad$ vertical $\approx 38.57 \mathrm{ft} / \mathrm{s}$

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Video Transcript

Yes. And this problem, we are given the distance that something is thrown. And we're told that we need to find the vertical and horizontal components of this velocity vector. Well, the first thing that we should do is let's just draw picture things going to help us find the distances of the the last two vectors that we still need the vertical and the horizontal. So when we do that, we're going to get this right triangle In this coordinate plain, we'll call this triangle ABC, and we know that this angle of elevation was 40 degrees. So we'll call that data. We're going to call this vector A, which is our, um, horizontal are vertical vector will call that be, and then we know that this vector has a magnitude of 60 ft per second. So the magnitude of our vector r equals well, the magnitude of our VR distance. So that's gonna be 60 ft per second. And now we need to know A and B. So how are we going to find that? Well, they will be equal, tow our times the coastline of data and be with equal the R times, a sign of data again. We have a right triangle forming with our vector so we can do this so it's fine. A A would be equal to 60 times a co sign of 40 will get 49.96 ft per second. So that is our horizontal component of our velocity vector. And then for B will do 60 times a sign of 40 which is 38.57 ft per second. And that would be the vertical component of our horizontal of pardon me are vertical component of our velocity vector. I hope this probably helped you understand a little bit more about vectors and how we confined essentially the magnitude and distance of vectors that would make a right triangle.