Minimum Angle for Increasing Distance of Projectile?

[Hashiramasenju]

Homework Statement

whats the minimum angles to the vertical(theta) for a projectile(ball) to be realeased with speed v such that at any point of time the distance to the ball is increasing.

Homework Equations

Sh=vtsin(theta)
Sv=vtcos(theta)-0.5gt^2

The Attempt at a Solution

so i used pythagoras theorum to find the square of the distance to the projectile(R). thus d(R^2)/dt>0 because the distance to the ball from the origin(start point) must be increasing. But the algebra gets really nasty. Is there any alternative way or is there any error in my method
Thanks

 

[Simon Bridge]

No, that's pretty much it. You probably need a table of trig substitutions.
Don't forget you only need dR/dt > 0 for 0 < t < T (time of flight)... in fact, if R decreases at all, which part of the trajectory will that happen?

There are other methods... ie you can guess some values and look for a pattern, you can try using your understanding of parabolas to narrow your choice, use graphing software... etc.

 

[Hashiramasenju]

No, that's pretty much it. You probably need a table of trig substitutions.
Don't forget you only need dR/dt > 0 for 0 < t < T (time of flight)... in fact, if R decreases at all, which part of the trajectory will that happen?

There are other methods... ie you can guess some values and look for a pattern, you can try using your understanding of parabolas to narrow your choice, use graphing software... etc.

Btw is the answer 45 deg

 

[Hashiramasenju]

No, that's pretty much it. You probably need a table of trig substitutions.
Don't forget you only need dR/dt > 0 for 0 < t < T (time of flight)... in fact, if R decreases at all, which part of the trajectory will that happen?

There are other methods... ie you can guess some values and look for a pattern, you can try using your understanding of parabolas to narrow your choice, use graphing software... etc.

So is the answer 45 deg and 35.26

 

[Simon Bridge]

The question only requires one number for the answer.
I don't know the correct answer off the top of my head.

 

[SammyS]

So is the answer 45 deg and 35.26

I get something a bit less than 20° .

 

[Hashiramasenju]

I get something a bit less than 20° .

How did you get that ?

 

[SammyS]

How did you get that ?

I did what you set up as follows.

The Attempt at a Solution

so i used pythagoras theorum to find the square of the distance to the projectile(R). thus d(R^2)/dt>0 because the distance to the ball from the origin(start point) must be increasing. But the algebra gets really nasty. Is there any alternative way or is there any error in my method
Thanks

R² = S_h² + S_v² is not as difficult to deal with as it might at first appear to be. R² is a degree 4 polynomial in t, but the constant term and the linear term are bot zero. Its derivative is a cubic polynomial in t, with a constant term of 0, so that one of the zeros is t=0, and you can use the quadratic formula for the other two zeros.

All that is needed is that the derivative is never zero, except perhaps at t = 0. Simply check the discriminant to determine those conditions.