Mathematical Proof That a 45 Degree Angle Launch is Best for Displacementby BasketDaN Tags: angle, degree, displacement, launch, mathematical, proof 

#1
Oct304, 08:14 PM

P: 96

Does anybody know how to mathematically prove that an object launched at a 45 degree angle will have the largest possible displacement?




#2
Oct304, 08:33 PM

Mentor
P: 7,291

Do the 2d solution to the free falling body problem, Obtain a solution for the range in terms of the initial velocity, this will include the initial angle, take a derivative with respect to x (the horizontal component), set equal to zero and solve for the angle.




#3
Oct304, 09:23 PM

Emeritus
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#4
Oct604, 01:37 AM

P: 91

Mathematical Proof That a 45 Degree Angle Launch is Best for Displacement
MAximum distance covered, d = (u^2sin[itex]2\theta[/itex])/g
u= velocity at time of projection for d to be maximum (u^2sin[itex]2\theta[/itex]) must have its maximum value for given speed, this implies that sin[itex]2\theta[/itex] must have maximum possible value that is 1, whem angle is 90 degrees so angle is obviously 45 degrees 



#5
Oct604, 10:47 PM

P: 21

Use the symmetry of the parabolic trajectory to double the horizontal (x) distance at the position of maximum height (y). The maximum height occurs when the vertical speed [itex]V_{y}[/itex] = 0. [tex]V_{0y} = V_0 sin\theta[/tex] As a function of time, t: [tex]V_{y} = V_0 sin\theta  gt[/tex] When [itex]V_{y} = 0[/itex]: [tex](1): V_0 sin\theta = gt[/tex] Now horizontal displacment: [tex]x = V_{0x}t = V_0 cos\theta t[/tex] [tex](2): t = \frac{x}{V_0 cos\theta}[/tex] Substituting for t from (2) in (1): [tex]V_0 sin\theta = \frac{gx}{V_0 cos\theta}[/tex] [tex]x = \frac{V_0^2 cos\theta sin\theta}{g}[/tex] The range is twice this distance so: [tex]x_{max} = \frac{2V_0^2 cos\theta sin\theta}{g}[/tex] Substituting the trigonometric identity:[itex] 2sin\theta cos\theta = sin2\theta[/itex] gives: [tex]x_{max} = \frac{V_0^2 sin2\theta}{g}[/tex] Since [itex]sin2\theta[/itex] is maximum at [itex]2\theta = \frac{\pi}{2}[/itex] then maximum horizontal displacement occurs at [itex] \theta = \frac{\pi}{4}[/itex] Now that you know this, you can give coaching tips to the punter on your college football team: kick the ball so that it has a 45 degree angle on takeoff. It also explains why my 5 wood goes farther than my drive (that and my slice). Calculex 


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