Max Angle of Projectile for Increasing Position Vector

[rattanjot14]

Homework Statement

1. Find the maximum angle of projection of a projectile such that its position vector from the origin to the subsequent position of the projectile is always increasing.

Homework Equations

The Attempt at a Solution

x(t)=v*cos(A)t, y(t)=v*sin(A)t-(1/2)gt^2. v is the initial velocity, A is the angle from the horizontal.
then x^2 +y^2 =k^2
We know that k^2 should always be increasing then we would differentiate it w.r.t to time and put it ≥ 0. But the problem is that i am getting two times for it. now what to do..?

 

[voko]

If you get some values of t, that means the magnitude of displacement will be increasing sometimes. You must find a condition when it always increases.

 

[rattanjot14]

i have a doubt when i differentiate it wrt to time i assume that theta is constant. can i do that..??

 

[voko]

Does the initial angle of projection change as time goes on?

 

[rattanjot14]

but we had to find the maximum angle of projection..

 

[voko]

And?

 

[rattanjot14]

only angle of projection for which the position vector always increases..

 

[voko]

You don't have to repeat the problem.

 

[rattanjot14]

so is it correct to assume it constant..?

 

[voko]

Answer #4.

 

[rattanjot14]

thnx answer is coming sin inverse 1/u

 

[voko]

I do not think this is correct. Show how you got that.

 

[rattanjot14]

r^2 = (u^2)t + (g^2)(t^4) - (usinθ)gt^3
differentiating it and putting it equal to 0 we get
(gt)^2 -3usinθgt +2u^2 =0
so t = (3usinθ ± √9u^2sin^2θ-8u^2)/20
and also this time ≤2usinθ/g (total time of flight)]
solving this i don't get the answer...i was getting that because of an error.
now what to do?

 

[voko]

Explain how you got (u^2)t in r^2 = (u^2)t + (g^2)(t^4) - (usinθ)gt^3.

 

[rattanjot14]

typing error it is . the actual it is

(u^2)t^2 + [(g^2)(t^4)/4] - (usinθ)gt^3.

 

[voko]

So how can one ensure that (gt)^2 -3usinθgt +2u^2 is ALWAYS greater than zero?

 

[rattanjot14]

if d is less than or equal to zero.

 

[rattanjot14]

answer is coming that 8/9≥(sinθ)^2

 

[rattanjot14]

is that correct?

 

[voko]

This is not yet the answer. But you are on the right track.

 

[rattanjot14]

maximum angle is sin inverse (2√2/3) ..is it correct now?

 

[voko]

That is correct.

I suggest that you pick an angle greater than this, and graph the resultant projectile motion, then mark on the diagram the part where the magnitude of displacement decreases.