CLASS7 - 1 CHAPTER 7 PROJECTILES 7.1 No Air Resistance We...

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1 CHAPTER 7 PROJECTILES 7.1 No Air Resistance We suppose that a particle is projected from a point O at the origin of a coordinate system, the y -axis being vertical and the x -axis directed along the ground. The particle is projected in the xy -plane, with initial speed V 0 at an angle α to the horizon. At any subsequent time in its motion its speed is V and the angle that its motion makes with the horizontal is ψ . The initial horizontal component if the velocity is V 0 cos α , and, in the absence of air resistance, this horizontal component remains constant throughout the motion. I shall also refer to this constant horizontal component of the velocity as u . I.e. u = V 0 cos α = constant throughout the motion. The initial vertical component of the velocity is V 0 sin α , but the vertical component of the motion is decelerated at a constant rate g . At a later time during the motion, the vertical component of the velocity is V sin ψ , which I shall also refer to as v . In the following, I write in the left hand column the horizontal component of the equation of motion and the first and second time integrals; in the right hand column I do the same for the vertical component. Horizontal. Vertical 0 = x g y - = 7.1.1 a , b α = = cos 0 V u x gt V y - α = = sin 0 v 7.1.2 a , b x V t = 0 cos α y V t gt = - 0 1 2 2 sin α 7.1.3 a , b The two equations 7.1.3 a , b are the parametric equations to the trajectory. In vector form, these two equations could be written as a single vector equation: r V g 0 = + t t 1 2 2 . 7.1.4 Note the + sign on the right hand side of equation 7.1.4. The vector g is directed downwards. The xy -equation to the trajectory is found by eliminating t between equations 7.1.3 a and 7.1.3 b to yield: y x gx V = - tan cos . α α 2 0 2 2 2 7.1.5
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2 Now, re-write this in the form ( 29 ( 29 . 2 y x x - = - Add to each side (half the coefficient of x ) 2 in order to "complete the square" on the left hand side, and, after some algebra, it will be found that the equation to the trajectory can be written as: ( 29 ( 29 , 4 2 B y a A x - - = - 7.1.6 where A V g V g = = 0 2 0 2 2 2 sin cos sin , α α α 7.1.7 B V g = 0 2 2 2 sin , α 7.1.8 and a V g = 0 2 2 2 cos . α 7.1.9 Having re-arranged equation 7.1.5 in the form 7.1.6, we see that the trajectory is a parabola whose vertex is at ( A , B ). The range on the horizontal plane is 2 A, or V g 0 2 2 sin . α The greatest range on the horizontal plane is obtained when sin 2 α = 1, or α = 45 o . The greatest range on the horizontal plane is therefore V g 0 2 / . The maximum height reached is B , or V g 0 2 2 2 sin . α The distance between vertex and focus is a , or V g 0 2 2 2 cos . α The focus is above ground if this is less than the maximum height, and below ground if it is greater than the maximum height. That is, the focus is above ground if cos sin . 2
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CLASS7 - 1 CHAPTER 7 PROJECTILES 7.1 No Air Resistance We...

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