A cannon tilted upward at 30° fires a cannonball with a speed of 100 m/s. What is the component of the cannonball's velocity parallel to the ground?

Video transcript

Hi, everyone in this particular practice problem, we are asked to calculate the ball's horizontal velocity component. When a volleyball Satter passes a ball to the OCI with a speed of 15 m per second. And there'll be, there'll be the V of equals to 15 m per second. And the velocity factor of the ball will make an angle of 20 degrees with the vertical. That will be the, the value that we have. And we were asked to calculate the ball's horizontal velocity. So I'm gonna start us off with um just making a diagram. I am going to skip the volleyball setter and this is going to be our ball right here. As always, the ball will have the uh velocity component. The velocity is going to be just like soap and this will be a V of 15 m per second. And I will draw a vertical and horizontal axis just to make it easy for us to actually see the projection and see the theta. So this is going to be our Y axis and this is going to be our ax axis or the vertical Y and the horizontal ax axis and the velocity factor of the ball will make an angle of 20 degrees with the vertical. So this angle right here with the vertical is going to be 20 degrees. So this will be theta equals 20 degrees. And the projection that's being asked is just the horizontal component, it's just the horizontal velocity. So I'm gonna draw that this is going to be VX which is the horizontal component of the velocity and that's what's being asked. OK. So this is the diagram of the system that we will have. And now we know that the horizontal velocity VX is going to be the projection of the velocity factor along the X axis. And we also know that the X and the Y axis will make a 90 degree angle just like. So, so we can actually find this angle right here or the angle between the velocity factor and the X axis component of the velocity, which is Phi well then equals to 90 degrees minus 20 degrees, which is the theta or Phi equals 90 degrees minus theta and that will come out to be 70 degrees. So using this Phi angle right here, we can find the horizontal component of the velocity PX which will equals to p multiplied by cosine of phi and the V is going to be 15 m per second and cosine of five is going to be cosine of 70 degrees just like. So and that will give us a VX value of 5.1 m per seconds. Another way of doing this is to buy a immediately using this data value here. And because uh doing that, then we can actually calculate the X by utilizing scent or the sign of data instead of the cosine because data is located opposite to the projection component that we want it. So doing this with the fee of 15 m per second and the the value of 20 degrees, it will give us the same value of VX, which will be 5.1 m per second. So 5.1 m per second is going to be the answer of or the management of the ball's horizontal velocity which will correspond to option A. So option A is going to be the answer to this particular practice problem. And if you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and they'll be all for this one. Thank you.