Register to reply 
Mathematical Proof That a 45 Degree Angle Launch is Best for Displacement 
Share this thread: 
#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

Emeritus
Sci Advisor
PF Gold
P: 7,337

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
Sci Advisor
PF Gold
P: 11,155




#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 


Register to reply 
Related Discussions  
Launch angle of projectile  Introductory Physics Homework  12  
Optimum Angle of Launch  Introductory Physics Homework  4  
ZERO Launch angle  Introductory Physics Homework  20  
Trouble with Proof of Optimal Launch Angle  Introductory Physics Homework  2  
Optimum Launch Angle  Introductory Physics Homework  1 