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

1. Oct 10, 2014

Dextrine

1. The problem statement, all variables and given/known data
"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?"

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

3. 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

2. Oct 10, 2014

RUber

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. Oct 11, 2014

Dextrine

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. Oct 11, 2014

Staff: Mentor

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:

(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. Oct 11, 2014

Staff: Mentor

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. Oct 11, 2014

RUber

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. Oct 12, 2014

Dextrine

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