1 CHAPTER 7 PROJECTILES. 7.1 No Air Resistance

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Dortha Warren
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1 CHAPTER 7 PROJECTILES 7 No Air Resistance We suppose that a particle is projected from a point O at the oriin of a coordinate system, the y-axis bein vertical and the x-axis directed alon the round The particle is projected in the xy-plane, with initial speed V at an anle α to the horizon At any subsequent time in its motion its speed is V and the anle that its motion makes with the horizontal is ψ The initial horizontal component if the velocity is V cos α, and, in the absence of air resistance, this horizontal component remains constant throuhout the motion I shall also refer to this constant horizontal component of the velocity as u Ie u V cos α constant throuhout the motion The initial vertical component of the velocity is V sin α, but the vertical component of the motion is decelerated at a constant rate At a later time durin the motion, the vertical component of the velocity is V sin ψ, which I shall also refer to as v In the followin, I write in the left hand column the horizontal component of the equation of motion and the first and second time interals; in the riht hand column I do the same for the vertical component Horizontal Vertical & x& & y 7a,b x& u cos α y& v V sin α t 7a,b V x V t cosα y V t sinα t 73a,b The two equations 73a,b are the parametric equations to the trajectory In vector form, these two equations could be written as a sinle vector equation: r V t + t 74 Note the + sin on the riht hand side of equation 74 The vector is directed downwards The xy-equation to the trajectory is found by eliminatin t between equations 73a and 73b to yield: x y x tan α V cos α 75

2 Now, re-write this in the form x ( ) x ( ) y Add to each side (half the coefficient of x) in order to "complete the square" on the left hand side, and, after some alebra, it will be found that the equation to the trajectory can be written as: ( x A) 4a( y B ), 76 where A V sinα cosα V sin α, 77 B V sin α, 78 and a V cos α 79 Havin re-arraned equation 75 in the form 76, we see that the trajectory is a parabola whose vertex is at (A, B) The rane on the horizontal plane is A, or V sin α The reatest rane on the horizontal plane is obtained when sin α, or α 45 o The reatest rane on the horizontal plane is therefore V / The maximum heiht reached is B, or V sin α The distance between vertex and focus is a, or V cos α The focus is above round if this is less than the maximum heiht, and below round if it is reater than the maximum heiht That is, the focus is above round if cos α < sin α That is to say, the focus is above round if α > 45 o and below round if α < 45 o The radius of curvature ρ anywhere alon the trajectory can be found usin the usual formula 3 ( + y' ) ρ At the top of the trajectory, y', so that ρ /y' Alternatively (in case one y" has forotten or is unfamiliar with the "usual formula"), we note that the speed at the top of the path is just equal to the (constant) horizontal component of the velocity V o cos α We can then equate the centripetal acceleration ( V cos α / ) to and hence obtain: ρ α ρ V cos 7

3 3 By subtractin this from our expression for the maximum heiht of the projectile, we find V ( 3cos α) that the heiht of the centre of curvature above the round is The centre of curvature is above round if α > 54 o 44' The rane r on a plane inclined at an anle θ to the horizontal can be found by substitutin x r cosθ and y r sin θ in the equation 75 to the trajectory This results, after some alebra, in V r [ sin( α θ) sin θ] 7 cos θ This is reatest when α θ 9 o ; ie when the anle of projection bisects the anle between the inclined plane and the vertical The maximum rane is V r 7 ( ) + sin θ This is the equation, in polar coordinates, of a parabola, and this parabola, when rotated about its vertical axis, describes a paraboloid, known as the paraboloid of safety It is the envelope of all possible trajectories with an initial speed V If a un is firin shells with initial speed V, or a lawn sprinkler is ejectin water at initial speed V, you are safe as lon as you are outside the paraboloid of safety Fiure VII shows trajectories for α, 4, 6, 8,,, 4 and 6 derees, and, as a dashed line, the paraboloid of safety Notice how the rane chanes with α and that it is reatest for α 45 o

4 Problem 4 A un projects a shell, in the absence of air resistance, at an initial anle α to the horizontal The speed of projection varies with anle of projection and is iven by Initial speed V cos α Show that, in order to achieve the reatest rane on the horizontal plane, the shell should be projected at an anle to the horizontal whose cosine c is iven by the solution of the equation 3 3c + c c Find the optimum anle to a precision of one arcminute 7 Air resistance proportional to the speed As in the previous section, I shall write the x-component of the equation of motion, and of the first and second time interals, in the left hand column, and the y-component in the riht-hand column The x-component of the air resistance per unit mass is γ x& and the y-component is γ y& Here γ is the dampin constant, defined in Chapter 6, section 3 The x- and y-components of the initial velocity are, respectively, V cos α and V sin α It should be readily seen that the equations of motion and their time interals are as follows: Horizontal Vertical & x γx& & y γ y& 7a,b γt γt γt x& u V cos α e y v V α e ˆ sin v( e ) & 7a,b where v ˆ / γ x t x ( γ e ) y ( V )( e γt sin α+ vˆ ) vˆ t where x V cosα γ γ 73a,b (In case it is not "readily seen", for the horizontal motion refer to Chapter 6, section 3, especially equations 63, 633 and 635, and for the vertical motion refer to Chapter 6, section 3b, especially equations 634, 635 and 637) It will be seen that, as t, u, v -vˆ, x x The xy-equation to the trajectory is the t-eliminant of equations 63a and 63b After a small amount of alebra this is found to be:

5 y x 5 ( V sin α + vˆ ) V cos α vˆ + ln γ x x 74 This is illustrated in fiure VII for the numerical data iven on the next pae Deleted: Deleted: The rane on a horizontal plane is found by settin y, to obtain either ( x ) x Aln / x 75 Deleted: x Deleted: x / A or x x ( e ), 76 where vˆ V cos α A, γ ( V sin α + vˆ ) α V cos x and v ˆ / γ γ Deleted: Formatted: Font: Bookman Old Style, Lowered by 4 pt

6 Example Suppose V m s - 6 α 5 o 98 m s - γ 96 s - ( v ˆ 5 m s ) Then A m and x m Try to find the rane on the horizontal plane, usin either equation 75 or 76, to nine sinificant fiures Which equation works best? Newton-Raphson may fail with a stupid first uess - but it should not be difficult to make a fairly intellient first uess I should not tell you, but fiure VII was calculated usin the data of this example I make the answer m Here s a more difficult problem: It is well known that, in the absence of air resistance, the maximum rane on the horizontal plane is effected by choosin the initial launch elevation to be α 45 o What if there is air resistance, with dampin constant γ? What, then, should be the anle of launch to achieve the reatest rane on the horizontal plane? Given equation 76, x / A x x ( e ), for what value of α is x reatest? Formatted: Lowered by 5 pt Formatted: Lowered by 5 pt Equation 76, written in full, is x V cos α γ γ( V sin α + vˆ) x exp 77 vˆ Vcosα Formatted: Lowered by 7 pt Formatted: Indent: Left: " This can be written ( bsin α + ) x x a cosα exp, cos 78 a α where V V γv a and b We have to find for what value of α is x reatest It γ v ˆ seems a simple enouh problem, but at the moment I can t find a ood way of solvin it If anyone has a clue, let me know (jtatum@uvicca) In the meantime, the best I can offer is, for our particular numerical example, to calculate the rane, x, for several values of α and see where it oes throuh a maximum For our particular numerical example, a m and Formatted: Tabs: 3", Left Formatted: Lowered by 5 pt Formatted: Lowered by 4 pt Formatted: Lowered by 4 pt

7 7 b 4 Here is a raph of rane versus launch anle, for an initial speed of m s A launch anle of about 3º 59 ives a rane of about m For a iven γ and, the optimum launch anle depends on the launch speed V Is this intuitively obvious? 8464 Formatted: Centered 8463 horizontal rane x Launch anle α derees 73 Air resistance proportional to the square of the speed Notation: V is the velocity, V is the speed The horizontal and vertical components of the velocity are, respectively, u x& V cosψ and v y& V sin ψ Here ψ is the anle that the instantaneous velocity V makes with the horizontal The resistive force per unit mass is kv The horizontal and vertical components of the resistive force per unit mass are kv cos ψ and kv sin ψ respectively The launch speed is V and the launch anle (ie the initial value of ψ) is α Distance travelled from the launch point, measured alon the trajectory, is s, and speed V s& The equations of motion are: Formatted: Space Before: pt, Tabs: 3", Left Deleted: Horizontal: & x& kv cos ψ 73 Vertical: & y& kv sin ψ 73

8 8 These cannot be interated as conveniently as in the previous cases, but we can et a simple relation between the horizontal component u of the speed and the intrinsic coordinate s Thus, when we make use of & x u&, V s& and V cos ψ u, equation 73 takes the form u & kus & 733 Interation, with initial condition u V cosθ, yields u V cos α e k s 734 We can also obtain an exact explicit intrinsic equation to the trajectory by consideration of the normal equation of motion The intrinsic equation to any curve is a relation between the intrinsic coordinates (s, ψ) The rate at which the slope anle ψ chanes as you move alon the curve, ie dψ/ds, is called the curvature at a point alon the curve If the slope is increasin with s, the curvature is positive The reciprocal of the curvature at a point, ds/dψ, is the radius of curvature at the point, denoted here by ρ The normal equation of motion is the equation F ma applied in a direction normal to the curve The acceleration appropriate here is the centripetal acceleration V /ρ or V dψ/ds In a direction normal to the motion, the air resistance has no component, and ravity has a component cos θ (It is minus because the curvature is clearly neative) The normal equation of motion is therefore V d ds ψ cos θ 735 But ks u V cos α e V 736 cos ψ cos ψ k s dψ 3 Therefore V cos α e cos ψ 737 ds Separate the variables, and interate, with appropriate initial conditions: 3 s ks sec ψ dψ e ds 738 V cos ψ α α From here it is ood interation practice to show that the intrinsic equation is

9 9 secψ + tan ψ k s secψ tan ψ secα tan α + ln ( e ) 739 secα + tan α kv cos α This equation is of the form s ( sec ψ + tan ψ) A Be k sec ψ tan ψ + ln 73 While it would be straihtforward now to compute s as a function of ψ and hence to plot a raph of s versus ψ, we really want to show y as a function of x, and x and y as a function of time I am indebted to Dario Bruni of Italy for the followin analysis Let (x, y ) be a point on the trajectory When the projectile moves a short distance s, the new coordinates will be (x, y ), where x 73 x + s cos ψ and y y + s sin ψ, 73 provided that s is taken to be sufficiently small that the path between the two points is approximately a straiht line The calculation starts with x y and ψ α At each stae of the calculation, the new value of ψ can be calculated from equation 73 This can be done easily, for example, by Newton-Raphson iteration, since the derivative of the left hand side of the equation with respect to ψ is just sec 3 ψ Thus, with a sufficiently small interval s, the shape of the trajectory can be built up point by point While this ives us the shape of the trajectory, it tells us nothin about the time To do this, we can write the equations of motion, equations 73 and 73 in the forms & x & & + & kx x y 733 and & y ky& x& + y& 734 Let (x, y ) be a point on the trajectory After a short time t, the new coordinates will be (x, y ), where x & & 735 x + x t + x ( t) and y y + y& t + & y ( ), 736 t provided that t is taken to be sufficiently small that the acceleration between the two instants of time is approximately constant Also, the new velocity components are iven by

10 x& x& + & x t 737 & and y& y& + & y 738 t The calculation starts with x & V cosα, y& V sin α, & x& kv cos α, & y& kv sin α, and after each increment t the new coordinates and velocity and acceleration components are calculated The results of Sr Bruni s calculations are shown in fiure VII3 for o k 77 m, V 95ms, α 6, 98ms 7 FIGURE VII3 6 5 y [m] x [m] Plotted with step by step method from intrinsic equation with s 5 m Horizontal rane 79 m; maximum heiht 64 m Total fliht duration 7 seconds The time taken to reach the maximum heiht is 8 seconds, so the descent time is loner than the ascent time

11 An alternative approach has been iven by Ambrose Okune, of Uanda In Okune s analysis, he obtains explicit expressions for t, x and y in terms of the anle ψ (In equation 73 we already have a relation between s and ψ) We start with equation 73, the horizontal equation of motion Now & x u&, V u + v, and V cos ψ u, so that & x& kv cos ψ kv V cos ψ 739 u& ku u + v 73 Similarly, equation 73, the vertical equation of motion, is & y& kv sin ψ kv V sin ψ, 73 and, with & y v&, V u + v and V sin ψ v, this becomes v & kv u + v 73 v& dv v Now + u& du u ku u + v 733 Also v u tan ψ, so that dv du tan ψ + dψ u sec ψ du 734 On comparison of equations 733 and 734, we see that ku u + v u sec ψ dψ 735 Upon substitution of v u tan ψ, this becomes ku 3 dψ sec 3 ψ 736 du and hence u 3 du sec 3 ψ d ψ k 737

12 Upon interation, we obtain ln(sec + ψ + tan ψ) + secψ tan ψ A + ln(secα + tan α) + secα tan α ku ku 738 From this, we obtain u, 739 k A ln(secψ + tan ψ) secψ tan ψ and hence tan ψ v 733 k A ln(secψ + tan ψ) secψ tan ψ Thus we now have the velocity components explicitly in terms of the anle ψ For simplicity, let us write λ A ln(secψ + tan ψ) secψ tan ψ 733 Then the equations for the velocity components are u 733 k λ tan ψ and v 7333 k λ o In the limit, as u, ψ 9, y, the motion approaches a vertical asymptote As tan ψ ψ 9 o, λ secψ tan ψ, and hence Lim o Thus the limitin value of the ψ 9 λ vertical component of the velocity is This arees precisely with what one would expect k for a body fallin vertically at terminal speed, with resistance proportional to the square of the speed (see equation 645) We now aim to find an expression relatin ψ to t, which we do by notin that

13 3 dψ dt du dt du dψ du dt du dλ dλ dψ 7334 The derivative du / dt can be found from the horizontal equation of motion & x& kv cos ψ, which can be written (because u V cos ψ and & x u& ) as u& ku secψ Then, makin use of equation 733, we obtain du secψ 7335 dt λ The derivative du / dλ can be found from equation 733 and is du dλ k 3/ λ 7336 The derivative d λ / dψ can be found from equation 733 and is dλ dψ sec 3 ψ 7337 Thus the relation we seek is dψ dt k λ cos ψ 7338 If the initial motion of the projectile at time zero makes an anle α with the horizontal, then interation of equation 7338 ives the followin expression for the subsequent time t when the motion makes an anle ψ with the horizontal dx dx dψ Also u With u and dt dψ dt we obtain dx dψ kλ α dψ t 7339 ψ k λ cos ψ dψ iven respectively by equations 733 and 7338, dt cos ψ from which we can calculate x as a function of ψ:, 734

14 4 x α dψ ψ k λ cos ψ 734 dy dy dψ Further, v dt dψ dt 7338, we obtain With v and dψ iven respectively by equations 7333 and dt dy tan ψ dψ kλ cos ψ, 734 from which we can calculate y as a function of ψ: tan α ψ dψ y 7343 ψ k λ cos ψ Equations 7339, 734 and 7343 enable us to calculate t, x and y as a function of ψ, and hence to calculate any one of them in terms of any of the others In each case a numerical interation is required, such as by Simpson s rule or by Gaussian quadrature, or other interation alorithm, and, as is always the case, sufficient points must be sampled to obtain adequate precision Numerical interation of these equations, usin the data of Dario Bruno s example above, produced the same x: y trajectory as calculated for fiure VII3 by Bruno, and the x : t and y : t relations shown in fiure VII4 I am reatly indebted to Dario Bruni and to Ambrose Okune for their interestin and instructive contributions to this section an inspirational example of international scientific cooperation between, Italy, Uanda and Canada!

15 5 8 FIGURE VII4 7 x 6 Distance, metres y Time, seconds

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