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Trajectory
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==Examples== === Uniform gravity, neither drag nor wind=== [[File:Inclinedthrow.gif|thumb|400px|Trajectories of a mass thrown at an angle of 70Β°:<br> {{color box|black}} without [[Drag (physics)|drag]]<br> {{color box|blue}} with [[Stokes' law|Stokes drag]]<br> {{color box|green}} with [[Newtonian fluid|Newton drag]]]] The ideal case of motion of a projectile in a uniform gravitational field in the absence of other forces (such as air drag) was first investigated by [[Galileo Galilei]]. To neglect the action of the atmosphere in shaping a trajectory would have been considered a futile hypothesis by practical-minded investigators all through the [[Middle Ages]] in [[Europe]]. Nevertheless, by anticipating the existence of the [[vacuum]], later to be demonstrated on [[Earth]] by his collaborator [[Evangelista Torricelli]]{{Citation needed|date=March 2009}}, Galileo was able to initiate the future science of [[mechanics]].{{Citation needed|date=March 2009}} In a near vacuum, as it turns out for instance on the [[Moon]], his simplified parabolic trajectory proves essentially correct. In the analysis that follows, we derive the equation of motion of a projectile as measured from an [[Inertial frame of reference|inertial frame]] at rest with respect to the ground. Associated with the frame is a right-hand coordinate system with its origin at the point of launch of the projectile. The <math>x</math>-axis is tangent to the ground, and the <math>y</math>axis is perpendicular to it ( parallel to the gravitational field lines ). Let <math>g</math> be the [[standard gravity|acceleration of gravity]]. Relative to the flat terrain, let the initial horizontal speed be <math>v_h = v \cos(\theta)</math> and the initial vertical speed be <math>v_v = v \sin(\theta)</math>. It will also be shown that the [[range of a projectile|range]] is <math>2v_h v_v/g</math>, and the maximum altitude is <math>v_v^2/2g</math>. The maximum range for a given initial speed <math>v</math> is obtained when <math>v_h=v_v</math>, i.e. the initial angle is 45<math>^\circ</math>. This range is <math>v^2/g</math>, and the maximum altitude at the maximum range is <math>v^2/(4g)</math>. ====Derivation of the equation of motion==== Assume the motion of the projectile is being measured from a [[free fall]] frame which happens to be at (''x'',''y'') = (0,0) at ''t'' = 0. The equation of motion of the projectile in this frame (by the [[equivalence principle]]) would be <math>y = x \tan(\theta)</math>. The co-ordinates of this free-fall frame, with respect to our inertial frame would be <math>y = - gt^2/2</math>. That is, <math>y = - g(x/v_h)^2/2</math>. Now translating back to the inertial frame the co-ordinates of the projectile becomes <math>y = x \tan(\theta)- g(x/v_h)^2/2</math> That is: : <math>y=-{g\sec^2\theta\over 2v_0^2}x^2+x\tan\theta,</math> (where ''v''<sub>0</sub> is the initial velocity, <math>\theta</math> is the angle of elevation, and ''g'' is the acceleration due to gravity). ====Range and height==== [[Image:Ideal projectile motion for different angles.svg|thumb|350px|Trajectories of projectiles launched at different elevation angles but the same speed of 10 m/s in a vacuum and uniform downward gravity field of 10 m/s<sup>2</sup>. Points are at 0.05 s intervals and length of their tails is linearly proportional to their speed. ''t'' = time from launch, ''T'' = time of flight, ''R'' = range and ''H'' = highest point of trajectory (indicated with arrows).]] The '''range''', ''R'', is the greatest distance the object travels along the [[x-axis]] in the I sector. The '''initial velocity''', ''v<sub>i</sub>'', is the speed at which said object is launched from the point of origin. The '''initial angle''', ''ΞΈ<sub>i</sub>'', is the angle at which said object is released. The ''g'' is the respective gravitational pull on the object within a null-medium. : <math>R={v_i^2\sin2\theta_i\over g}</math> The '''height''', ''h'', is the greatest parabolic height said object reaches within its trajectory : <math>h={v_i^2\sin^2\theta_i\over 2g}</math> ====Angle of elevation==== [[File:Selomie Melkie - Forensics Final Project (5).jpg|thumb|An example showing how to calculate bullet trajectory]] In terms of angle of elevation <math>\theta</math> and initial speed <math>v</math>: :<math>v_h=v \cos \theta,\quad v_v=v \sin \theta \;</math> giving the range as :<math>R= 2 v^2 \cos(\theta) \sin(\theta) / g = v^2 \sin(2\theta) / g\,.</math> This equation can be rearranged to find the angle for a required range : <math> \theta = \frac 1 2 \sin^{-1} \left( \frac{g R}{ v^2 } \right) </math> (Equation II: angle of projectile launch) Note that the [[sine]] function is such that there are two solutions for <math>\theta</math> for a given range <math>d_h</math>. The angle <math>\theta</math> giving the maximum range can be found by considering the derivative or <math>R</math> with respect to <math>\theta</math> and setting it to zero. :<math>{\mathrm{d}R\over \mathrm{d}\theta}={2v^2\over g} \cos(2\theta)=0</math> which has a nontrivial solution at <math>2\theta=\pi/2=90^\circ</math>, or <math>\theta=45^\circ</math>. The maximum range is then <math>R_{\max} = v^2/g\,</math>. At this angle <math>\sin(\pi/2)=1</math>, so the maximum height obtained is <math>{v^2 \over 4g}</math>. To find the angle giving the maximum height for a given speed calculate the derivative of the maximum height <math>H=v^2 \sin^2(\theta) /(2g)</math> with respect to <math>\theta</math>, that is <math>{\mathrm{d}H\over \mathrm{d}\theta}=v^2 2\cos(\theta)\sin(\theta) /(2g)</math> which is zero when <math>\theta=\pi/2=90^\circ</math>. So the maximum height <math>H_\mathrm{max}={v^2\over 2g}</math> is obtained when the projectile is fired straight up. ===Orbiting objects=== If instead of a uniform downwards gravitational force we consider two bodies [[orbit]]ing with the mutual gravitation between them, we obtain [[Kepler's laws of planetary motion]]. The derivation of these was one of the major works of [[Isaac Newton]] and provided much of the motivation for the development of [[differential calculus]].
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