Solving for maximum angle theta so that the distance of a projectile fired is constantly increasingn

In summary: I set up the systemdr/dt = r(t)+v(t)v(t)=-r(t)^2I solved for t_final and got it to be t_final=4.905+0.5t+0.525t^2
  • #1
Dextrine
102
7

Homework Statement


"What is the maximum angle above the horizontal with which a projectile can be
fired so that its distance from its point of firing (on earth) is always increasing?"
[/B]

Homework Equations


r^2=t^2 v^2 Cos[x]^2 + (-4.905 t^2 + t v Sin[x])^2

The Attempt at a Solution


I have tried setting dr/dt > 0 so that the distance is always increasing as well as I have also tried setting
dr^2/dt > 0. However, I don't know how to solve the equation after getting to this point(mathematica doesn't help either as it says there is no method for solving that equation with Solve[]). Or even if that is what I'm supposed to do
 
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  • #2
What did your work look like when you set dr/dt>0. This sounds like the right approach.
Based on what you have above, carrying out the squaring of the polynomial and combining the (sin^2 + cos^2 ) terms should give the form below.
## r^2 = t^2 v^2 \cos^2 x + 4.905^2t^4+t^2v^2\sin^2 x - 9.810t^3 v \sin x ##
##r^2=t^2 v^2+ 4.905^2t^4- 9.810t^3 v \sin x##
 
  • #3
That is the same that I got. The derivative I get is 2v^2t+3vgt^2sin(theta)+g^2t^3 which I set equal to 0 then solve for theta and get theta=arcsin[(-g^2t^3-2v^2t)\(3vgt^2)]
 
  • #4
If you plot an example ##r^2(t)## for the case where your angle x is bound to be too large then you'll get a curve that looks something like this:

Fig1.gif

(For the plot I assumed an initial velocity of 1 m/s and chose a launch angle of 73°. The plot covers the trajectory from launch to time of landing assuming level ground)

What needs to happen in order to make that downslope portion (where the distance-squared is decreasing) go away?
 
  • #5
Don't plug in a numerical value for g, it just makes the numbers messy.
Apart from that, you can follow gneill's idea or see at which point the definition of theta found in post 3 does not work any more.
 
  • #6
RUber said:
##r^2=t^2 v^2+ 4.905^2t^4- 9.810t^3 v \sin x##
Taking the derivative with respect to time will give t( quadratic ).
When you are want a function to always be positive, you are saying that there will be no real roots, or that the stuff inside the square root in the quadratic equation will always be less than or equal to zero.
This should allow you to get a numerical value for x that does not depend on v.
 
  • #7
Just in case anyone was wondering, I tried a different method and was able to get an answer. starting with that statement that if
| r(t_final) | > | r(t) | then dr/dt>0
 

1. What is the maximum angle theta for a projectile to have a constantly increasing distance?

The maximum angle theta for a projectile to have a constantly increasing distance is 45 degrees. This angle, also known as the optimal angle, allows the projectile to travel the furthest distance due to a balance between horizontal and vertical components of velocity.

2. How do you calculate the maximum angle theta for a projectile?

The maximum angle theta can be calculated using the equation:
theta = arctan((v2 +/- sqrt(v4 - g(gx2 + 2yv2))) / gx), where v is initial velocity, g is the acceleration due to gravity, and x and y are the horizontal and vertical distances, respectively.

3. Can the maximum angle theta be greater than 45 degrees?

No, the maximum angle theta cannot be greater than 45 degrees. This is because at angles greater than 45 degrees, the vertical component of velocity becomes greater than the horizontal component, resulting in a shorter distance traveled.

4. What factors affect the maximum angle theta for a projectile?

The initial velocity, acceleration due to gravity, and horizontal and vertical distances all affect the maximum angle theta for a projectile. A higher initial velocity and longer horizontal distance will result in a larger maximum angle theta, while a higher acceleration due to gravity and shorter vertical distance will result in a smaller maximum angle theta.

5. Why is it important to find the maximum angle theta for a projectile?

Finding the maximum angle theta for a projectile is important in determining the optimal angle for maximum distance traveled. This can be useful in various applications, such as sports like baseball or in engineering projects involving projectile motion.

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