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Mathematical Proof That a 45 Degree Angle Launch is Best for Displacement

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BasketDaN
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Oct3-04, 08:14 PM
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Does anybody know how to mathematically prove that an object launched at a 45 degree angle will have the largest possible displacement?
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Integral
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Oct3-04, 08:33 PM
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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.
Gokul43201
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Oct3-04, 09:23 PM
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Quote Quote by BasketDaN
Does anybody know how to mathematically prove that an object launched at a 45 degree angle will have the largest possible displacement?
Write down the expression for the range, in terms of [itex]2\theta[/itex]. That will make it obvious.

aekanshchumber
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Oct6-04, 01:37 AM
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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
Calculex
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Oct6-04, 10:47 PM
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Quote Quote by BasketDaN
Does anybody know how to mathematically prove that an object launched at a 45 degree angle will have the largest possible displacement?
I assume the object is landing at the same level as it is launched. Otherwise, it is not correct that a 45 degree angle gives the largest horizontal displacement.

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 take-off. It also explains why my 5 wood goes farther than my drive (that and my slice).

Calculex
Gokul43201
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Oct6-04, 11:04 PM
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Quote Quote by Calculex
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 take-off.

Calculex
But range is not the only thing a punter worries about. There's hang time too, especially if you have a short punt and don't want a touch back.


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