Optimun angle of launch

  • #1

Main Question or Discussion Point

I've been trying to determine the optimun angle of launch 1,5 metres above the ground by pooceeding from the basic equations for throwing motions:

x=vcosθ*t
y=y0+vsinθ*t-gt^2/2, y0=1,5m, g=9,8m/s^2

What I've done so far is expressing t as x/vcosθ. Insertion in the equation for y gives:
y=y0+xvsinθ/vcosθ-(g/2)(x/vcosθ)^2

y equals 0 when the object hits the ground. This means:
0=y0+xvsinθ/vcosθ-(g/2)(x/vcosθ)^2

I've then expressed x as a function of θ by solving the quadratic equation above.
It's desirable to derive the function and then solve the equation x'(θ)=0 to find what angle that gives the maximum range 1,5 metres above the ground.
I have however been unsuccesful so far and would be very grateful for any advice regarding how to derive the function and solving the following equation. I'm not that advanced in physics and would therefore apreciate thorough explanations.

Please help me out here, because I'm totally lost.
 

Answers and Replies

  • #2
154
0
Don't really understand what you're trying to determine. Are you trying to figure out all the trajectories that have a maximum altitude of 1.5 meters?
 
  • #3
I'm trying to determine the angle that will result in the theoretical maximal range 1,5 meters above the x-axis in a coordinate system. Everyone knows that the optimal angle of launch is 45 degrees when launching from origo. A vertical displacement results in a slightly smaller angle.
 
  • #4
106
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Would raising something 1.5 meters off the ground really stop the optimal angle being 45 degrees?
 
  • #5
993
13
Would raising something 1.5 meters off the ground really stop the optimal angle being 45 degrees?
I do not think so.
 
  • #6
A vertical displacement does indeed change the optimal angle. Imagine that you're standing on a mountain 1000m above the ground level and are about to throw something. Due to the position the horizontal vector vcosθ needs to be much bigger than if the same situation would have occured on the ground level. To recieve the maximum range that the object can be throwed you'll have to calculate this new optimal angle. This angle will of course be much smaller than 45 degrees. Otherwise too much énergy would be wasted on the vertical movement.
 
  • #7
390
1
We want to maximize [itex] x = V_0 \cos (\theta) t [/itex] where [itex] V_0 [/itex] is the initial velocity [itex] \theta [/itex] is the launch angle above the horizontal and [itex] t [/itex] is obviously the time of flight. Since we also know that
[tex] y = y_0 + V_0 \sin (\theta)t + \frac{1}{2}gt^2 [/tex]
With [itex] y_0 [/itex] being the initial vertical displacement; we can then express t in terms of this displacement, the initial angle and the initial velocity
[tex] t = \frac{\sqrt{V{_0}^2 \sin^2 \theta - 2gy_0}-V_0 \sin \theta}{g} [/tex]
So our equation we want to minimize now becomes
[tex] x = V_0 \cos \theta \frac{\sqrt{V{_0}^2 \sin^2 \theta - 2gy_0}-V_0 \sin \theta}{g} [/tex]
Now, to maximize we differentiate with respect to [itex] \theta [/itex] while treating [itex] V_0 [/itex] constant and taking the answer in the first quadrant we get
[tex]\theta = \cos ^{-1}- \sqrt{\frac{2y_0g - V{_0}^2}{2y_0g - 2V{_0}^2}} [/tex]
Which (should, it's getting a little late over here) gives the optimal angle of launch, neglecting air resistance, of a projectile initially [itex] y_0 [/itex] units above a plane for maximum distance :)
Plugging in [itex] y_0 = 0 [/itex] we get the usual [itex] \frac{\pi}{4} [/itex] answer we should :)
If anybody can spot any errors please feel free to point them out, like I said things are getting a bit sloppy over my side of the world
:)
 
  • #8
That is indeed a very interesting method. However, when I plug 1,5 at y0 I get something bigger than 1 under the square root. This means that the expression isn't defined since -1<cos^-1<1.
 
  • #9
Born2bwire
Science Advisor
Gold Member
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That is indeed a very interesting method. However, when I plug 1,5 at y0 I get something bigger than 1 under the square root. This means that the expression isn't defined since -1<cos^-1<1.
He forgot that acceleration is in the negative y direction. Working it out myself I get

[tex]\theta = \sin^{-1}\left( \frac{V_0}{\sqrt{2 \left(V_0^{\ 2}+gy_0 \right)}} \right) [/tex]
 
Last edited:
  • #10
1,945
242
He forgot that acceleration is in the negative y direction. Working it out myself I get

[tex]\theta = \sin^{-1}\left( \frac{V_0 \sqrt{2V_0^2 m^2 + 2gmy_0} }{2mV_0^2+2gy_0} \right) [/tex]
This can't be right. The optimum angle doesn't depend on mass.
 
  • #11
Born2bwire
Science Advisor
Gold Member
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This can't be right. The optimum angle doesn't depend on mass.
Yeah I don't know why I put that in there, Matlab can be a bit lazy about simplifying stuff. There we go.
 
  • #12
The initial velocity doesn't effect the optimal angle for a specific vertical displacement. When I use the formula above and plug in different values at v0 and 1,5 at yo I get different results. Logically, the equation for θ shouldn't involve v0 at all.
 
Last edited:
  • #13
All my attempts to determine the optimal angle of launch with a vertical displacement of 1,5 have been unsuccessful. Is it impossible to calculate?
 
  • #14
Born2bwire
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All my attempts to determine the optimal angle of launch with a vertical displacement of 1,5 have been unsuccessful. Is it impossible to calculate?
What's wrong with the answer we gave you above?
 
  • #15
1,945
242
What's wrong with the answer we gave you above?
He thinks that

The initial velocity doesn't effect the optimal angle for a specific vertical displacement. When I use the formula above and plug in different values at v0 and 1,5 at yo I get different results. Logically, the equation for θ shouldn't involve v0 at all.
But that is simply wrong. With a very large v0 the 1.5 m altitude matters little and the launch angle is very close to 45 degrees, with a very small v0, the time to reach the ground will be nearly independent of v0, and you should launch nearly horizontally.
 

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