During a baseball game a baseball is struck at ground level...

During a baseball game, a baseball is struck at ground level by a batter. The ball leaves the baseball bat with an initial speed v0 = 38 m/s at an angle θ = 28° above horizontal. Let the origin of the Cartesian coordinate system be the ball's position the instant it leaves the bat. Air resistance may be ignored throughout this problem. A.) Express the magnitude of the ball’s initial horizontal velocity v0x in terms of v0 and θ. B.) Express the magnitude of the ball’s initial vertical velocity v0y in terms of v0 and θ. C.) Find the ball’s maximum vertical height hmax in meters above the ground. D.) Enter an expression in terms of v0, θ, and g for the time tmax it takes the ball to travel to its maximum vertical height. E.) Calculate the horizontal distance xmax in meters that the ball has traveled when it returns to ground level.

Transcript

00:01 Okay, so we have a ball being hit at a velocity, initial velocity of 38 meters per second, and that's at an angle of 28 degrees from the horizontal.

00:11 And we want to know a few things about this situation.

00:16 So, first of all, we'll figure out what the x component, the horizontal component of the velocity is initially.

00:28 So we've got v initial in the x direction, and that's just going to be the initial, that 38 meters per second, times cosine of theta.

00:47 Right.

00:48 So this is just the horizontal component of basically this triangle here.

00:56 So just that horizontal component is going to be v .0 cosine of theta, where v .0 is kind of the hypotenuse of that triangle.

01:10 Okay, and then if we just go ahead and plug in what we have here, so that's 38 meters per second times cosine of 28 degrees, that gives us.

01:31 34 meters per second.

01:36 And we can just do the same thing with the vertical component.

01:42 So v .0 in the y direction, it's going to be v .0 times sine of theta equals 38 meters per second times sine of 28 degrees.

02:00 And that is equal to.

02:03 To about 18 meters per second in the vertical direction.

02:10 Okay, and then we want to know our height h in this problem.

02:18 So this is going to be basically the height from zero.

02:27 So let's go ahead and figure that out.

02:32 So we can use this equation v final equals v initial, plus acceleration times time.

02:44 So the initial in the y direction.

02:48 Okay, and so at y equal to h max, at that point, the velocity is going to be zero because you have the velocity decreasing, well, okay, the velocity in the y direction.

03:10 So the velocity is decreasing up until you hit that h max, and then it stops for just a second, and then it starts going in the opposite direction.

03:24 So at that point, our v is zero.

03:28 So we can use that in these equations to solve first for the time, and then we'll find the height.

03:36 So we'll use this equation to find the time first.

03:40 It takes to get to the maximum height.

03:44 All right...

Question

Please give Ace some feedback