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MMH On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 3.5 times as far as he would have on earth, given the same initial velocities on both planets. The ball islaunched at a speed of 45 m/s at an angle of 29 above the horizontal. When the ball lands, it is at the same level as the tee. On the distant planet, what are (a) the maximum height and (b) the range of the ball?

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Physics 101 Mechanics

Chapter 3

Kinematics in Two Dimensions

Motion in 2d or 3d

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Washington

Simon Fraser University

Lectures

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2D kinematics is the study of the movement of an object in two dimensions, usually in a Cartesian coordinate system. The study of the movement of an object in only one dimension is called 1D kinematics. The study of the movement of an object in three dimensions is called 3D kinematics.

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A vector is a mathematical entity that has a magnitude (or length) and direction. The vector is represented by a line segment with a definite beginning, direction, and magnitude. Vectors are added by adding their respective components, and multiplied by a scalar (or a number) to scale the vector.

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So the question states that on Earth a ball is hit at 45 meters per second at a 29 degree angle. And we're told that on a distant planet the ball travels 3.5 times the range that it does on Earth. And we're trying to find the max height of the ball and its range on this distant planet. So the first thing we really need to do is find a relationship between the displacement of the ball and the acceleration on the planet. So to do this, um, we know that the range of the ball is going to be the velocity in the X direction. So it's gonna be the total magnitude of lost E time. Some co sign of an angle times sometime will be the range. So until tax and we want to substitute in for this time something that has acceleration in it. So we can use another formula which states that the the change in the white displacement is equal to the initial velocity times, time post, 1/2 times the acceleration times time squared. And we know that when the ball uh, since the ball starts on the ground and it ends on the ground. It's why displacement is going to be zero. We know that the initial velocity in the Y direction is just the total velocity times sine of the angle that it makes with the horizontal times time plus 1/2 time, some acceleration times, Times Square And so we can solve for T. And we'll find that tease equal to zero or T is equal to minus two times the velocity times sign of data all over the acceleration. And so we complete this time back in here and, well, this will give us our relationship that we need for the, uh, range and the acceleration. So get native to these squared signed they'd, uh, cu science data. Oh, her acceleration. And so, using this relationship, we know that on this distant planet, orders call it Delta X. So p, the range on this distant planet is equal to 3.5 times the range on Earth. So Delta X o. B. So if we plug in our Delta X values in here, we'll get the equation which states, um, negative to these squared time. Signed data co sign data over the acceleration on this this didn't plan. It is equal to negative seven a Times B squared sign data co sign data over the acceleration on Earth. And when we simplify, this will find that the two over the acceleration on a distant planet is equal to seven over the acceleration on Earth. Um, and once we simplify this once again, we'll find that the acceleration on the planet is equal to two sevens the acceleration on Earth, which is the same thing as to seventh times negative 9.8, which is equal to negative 2.8 meters per second squared. And this is the acceleration on the distant planet. And so now that we know this, we can calculate, um, everything we need Teoh. Fine, because the velocity is the same. The initial velocity is the same on the distant plane as well as the angle. The only thing that's different is just the acceleration. So to find the, um, maximum height of the the project oh, we can use the equation which states that the initial velocity squared is equal to the are sorry. The final velocity squared is equal to the initial velocity squared, plus two times the acceleration times, Delta, X or the displacement. So when the ball is at its maximum height, the velocity is going to be equal to zero in the vertical direction. We know that in the vertical direction, the initial velocity is equal to 45 times sign of 29 degrees. So this would be 45 time sign of 29 degrees squared, plus two tons Thea acceleration, which is negative. 2.8 that we calculated meters per second squared. And we're trying to find the height it reaches. So our guilt acts. So when we subtract, uh, negative 45 when we subtract 45 signed 29 degrees, squared on those sides and then divide by two times negative 2.8 we find that Delta X is equal to 85 meters. So that's the solution into the first part of the problem and the second part of the problem. As for the range of the projectile and to do this weaken juice, the equation that we solve for the are the equation we derived for the range originally, uh, this one. So let me write it out one more time, so Delta X is equal to negative two times the lost square time to sign Data Times Co. Signed data divided by the acceleration. And so when we plug in everything we know, so plug and negative, too times UH, 45 meters per second squared times sign of 29 degrees times Co sign of 29 degrees, all divided by a negative 2.8 meters per second squared. We get that. The total range is 613 0.32 meters, and that's the second answered.

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