At t = 0 a batter hits a baseball with an initial speed of 28 ...

Question

At t = 0 a batter hits a baseball with an initial speed of 28 m/s at a 55° angle to the horizontal. An outfielder is 85 m from the batter at t = 0 and, as seen from home plate, the line of sight to the outfielder makes a horizontal angle of 22° with the plane in which the ball moves (see Fig. 3–66). What speed and direction must the fielder take to catch the ball at the same height from which it was struck? Give the angle with respect to the outfielder's line of sight to home plate.
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Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A soccer player kicks a ball, launching it into the air at the moment, the clock starts ticking t equals zero. As shown, the ball has an initial velocity of 20 m per second directed at an upward angle of 45 degrees above the ground. Meanwhile, a goalkeeper stands ready positioned 50 m away from the point of the kick from the perspective of the kickoff point, the goalkeeper's direct line of sight to be the home spot forms a horizontal angle of 30 degrees with the plane of the ball's flight path, determine the speed and direction the goalkeeper must move to intercept the ball at the same height from which it was initially kicked. Give the angle with respect to the goalkeeper's line of sight to the kick point, hint, use law of signs and cosigns. Ok. So our end goal is to solve for two separate answers. First, we're asked to figure out what the speed and direction the goalkeeper must move in order to intercept the ball at the same height from which it was initially kicked. And then we're also asked to figure out what the angle is with respect to the goalkeeper's line of sight to the kick to the kick point. So those are the two answers that we're trying to solve for. And those will be our final answers that we're ultimately trying to solve for. Awesome. So let's quickly look at our diagram here that's provided to us by the prom itself. So we have a kind of like a shadow, I guess on the far left. So it's like a white stick figure which denotes our soccer player that's kicking a soccer ball. And as we can see are we have a solid black stick figure denoting the goalkeeper where they're initially standing and they're gonna move to this other location where their shadow is right now. So it's a little gray stick figure of the goalkeeper. So it's denoting, that's where the goalkeeper will be standing after he moves in order to intercept the ball. So he's standing at one point and somewhere on the field or in front of the goal and he has to move some direction, some distance, some angle in order to intercept the ball. Awesome. And as we could see, our soccer player will kick our soccer ball at 20 m per second and this is our initial velocity and it's at an angle of 45 degrees. And from where the ball is located to where the coal like the goalkeeper standing at one point and then he moves to another location. And when he moves to that other location, in order to intercept the ball, he has moved an angle of 30 degrees. As according to our diagram that's provided to us and from the ball to the goalkeeper, it's 50 m awesome. We're also given some multiple choice answers. Let's read them off to see what our final answer pair might be. And let's note that for the speed, it's all in units of meters per second and for the angle, it's all in units of degrees. So A is 7.5 and 45 B is 8.7 and 54 C is 9.2 and 60 D is 10 and 30. Awesome. So first off, let us make the following assumptions. Firstly, the ball's flight is only affected by its initial velocity and gravity. Secondly, the goalkeeper can move at a constant velocity. So now at this stage, we must create a more detailed diagram of this problem. So with that in mind, I put together a fancy diagram to help us better visualize what's going on. So as we could see, we have our soccer player kicking a soccer ball, which is denoted with a black dot which is the soccer ball and he strikes the ball and at an initial velocity of 20 m per second at an angle of 45 degrees. And from the ball, we have a dotted line going straight in the middle, denoting capital RR range. And as you can see, we have another little arch dotted line which denotes the ball as it's kicked up into the air. So it's gonna go so like projectile motion, get that in to start thinking about projectile motion. So the ball is gonna be kicked and it's gonna go up into the air where the goalkeeper is gonna move from where he's standing as he's black stick figure and move to where the gray stick figure is to intercept the ball. So the ball is gonna go kick woo and it's gonna go up in an art shape like a per parabolic shape of, of course, you know projectile motion, it's gonna go land here where the goalkeeper is gonna intercept it and that's our range value is where the goalkeeper is gonna move to intercept the ball. And from our range dotted line to the distance dotted line from where the ball and the goalkeeper are. The distance between them is 50 m. So it's an angle of 30 degrees and the distance that the goalkeeper has to move in order to intercept the ball is X. OK. Fantastic. So now we must solve for the value of R the range by recalling and using the following equation which states that are the range is equal to V zero squared multiplied by sine of two multiplied by theta all divided by G which denotes gravity. So now at this point, we can plug in all of our known variables to solve for R. So R is equal to our initial velocity, which V zero denotes the initial velocity. So it's 20.0 m per second, which you can just simply write 20 m per second. So meters per second multiplied by sign of two multiplied by. And our angles given to us as 45 degrees because that's the angle at which the soccer player strikes the ball divided by gravity, which G denotes gravity and the acceleration due to gravity is 9.8 m per second squared. So when we plug that into our calculator, we will get 40.816 m. Let me round to three decimal places. And as you can see if we quickly go through here, our seconds will cancel leaving us just with meters, which is perfect. Awesome. So with that in mind, oh And don't forget that this is squared. So that's where the second squared will cancel out. OK. So moving right along. So now we must solve for the distance that the goalkeeper needs to cover in order to reach the ball. And this value is X like we discussed in our diagram. So we can recall and use the law of cosines to solve for X. And let's note that the law of cosine states that C squared is equal to A squared plus B squared minus two multiplied by a multiplied by B multiplied by cosine of capital C. Thus, we can write that X squared is equal to 40.816 m squared plus 50 meters squared minus two, multiplied by 40.816 m multiplied by 50 m multiplied by cosine of. And this time we're gonna use 30 degrees because remember we're focusing on how far the angle at which the goalkeeper is moving in order to intercept the ball. OK. So we will find when we get X all by itself that X is equal to the square root of 631.177 when we round the three decimal places, meters squared, which is equal to when we plug that into our calculator, 25.123 which is approximately equal to 25 m when we round to the nearest whole number. So now at this point, we need to solve for the amount of time needed for the goalkeeper to intercept the ball. So we need to recall and use the following equation to solve for T the time. So T is equal to the range R divided by VX zero, which is equal to when we plug in our known variables. 40.816 m divided by 20 m per second multiplied by cosine of 45 degrees. Because VX zero is our X component of velocity. So we have to recall and use that equation and that's how we got our velocity multiplied by cosine of 45 degrees. So when we plug that into our calculator, we will get 2.886 seconds when we round the three decimal places. So now we must solve for the goalkeeper speed by recalling and using the following equation that V equals 25.123 m divided by 2.886 seconds, which is equal to 8.7 m per second when we round to one decimal place. And that is our first answer. Hooray, we did it. So now we must solve for the direction by finding out what the value for theta our angle is. So we need to recall and use the law of science which will help us write that are the range. So we'll take our, our range, divide it by sin of theta, which is equal to X divided by sine of 30 degrees, which we can simplify and write that sin of theta is equal to one half multiplied by 40.816 m, 8 1 6 m divided by 25.123 m. OK? Make sure we know that that's a parenthesis. So when we plug that into our calculator, we will get 0.8123 when we round to four decimal places. And finally, just to get sign all by itself, sorry to get theta, all by itself, we will take that theta is equal to inverse sign of 0.8123. So when you take inverse sign, that's key, that's how we get the all by itself. So when we take inverse sign of 0.8123 we will get 54 degrees when we round to the nearest whole number. And that is our final answer. So looking at our multiple choice answers, the correct answer has to be the letter B 8.7 m per second at 54 degrees. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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