Find the Angle for Maximum x Position in Free Fall Motion

[agro]

Consider a particle located at (x₀, y₀), having inital velocity v₀, and the angle of inclination of the velocity vector is [the]. The particle is subjected to a vertical acceleration of -g. It's equation of motion is...

x = x₀ + v₀*cos[the]*t
y = y₀ + v₀*sin[the]*t - 0.5*g*t²

If t>0, the particle will reach y = 0 when
t = {v₀*sin[the] + sqrt((v₀*sin[the])²+2*g*y₀)}/g

(found using the abc formula)

If we call the time t₁, then at that time the x position of the particle is (call it x₁)...

x₁ = x₀ + v₀*cos[the]*t₁

The question is: at initial height y₀, what angle will make x₁ maximum?

I found it very hard to make the equation simple enough to be able to solve this problem... Can anyone help me?

PS: in the case of y₀ = 0, the problem will be easy to solve, with the answer 45 degrees.

Thank you...

 

[HallsofIvy]

Replacing t1 in x1= x0+ v0 cos[theta] t1

with t1= (v0/g)sin[theta]+ (1/g)(v0^2 sin^2 [theta]+ (gyo/2))^(1/2)
(the positive time at which y= 0) gives

x1= x0+ (v0^2/g)sin[theta]cos[theta]+(v0/g)sin[theta](v0^2sin^2[theta]+ (gy0/2))

Differentiating,

dx1/d[theta]= (v0^2/g)(cos^2[theta]- sin^2[theta])+ (v0/g)cos[theta](v0sin^2[theta]+(gy0/2))^(1/2)+ vo^3/g sin^2[theta]cos[theta](v0^2sin^2[theta] + (gy0/2))^(-1/2) = 0 at maximum x1.

I'm with you! I don't see any "nice" way of solving that for [theta]. Do you have any reason to think that there should be?