Maximizing the distance of a projectile

[IDumb]

1) This is not a homework problem, just a question I came across and can't solve adequately.
2) Sorry if this is wrong area to post it.

Homework Statement

What angle would be best to hold your head at relative to ground to maximize the distance you can spit.

The Attempt at a Solution

I drew a diagram, horizontal axis was distance in the x direction, vertical axis was distance in y direction. I drew a vector for velocity, V₀ at an angle $$\theta$$ relative to ground. v_x and v_y represent the x and y components of this initial velocity vector. The initial velocity starts from a height of h.

I then divided it into 2 equations, the acceleration in the y and x directions.

The Y direction:
a_y = -9.8 (m/s²)
integrate twice to obtain position function:
p_y(t) = -4.9t² + v_yt + h

Next, I found the time, t1, that the projectile (saying spit sounds weirder) would reach ground, at y = 0.

p_y(t1) = 0
using the quadratic equation I obtained 2 results, 1 of which had to be negative. So the other one is the time we want.

t1 = V_y/9.8 + (1/9.8)([tex]\sqrt{v_y² + 19.6h})

OK, so at this time, we will have y = 0 and x = D, the farthest distance the projectile will reach.

The x equation:
a_x = 0 --> v_x(t) = v_x --> p_x(t) = v_xt

replace t with the value found for t1, because that will give us an equation for D, the maximum distance.
v_x(t1) = D = \frac{v_xv_y}{9.8} + (1/9.8)(v_x\sqrt{v_y² + 19.6h)

 

[gneill]

Okay, so how are v_x and v_y related to the launch angle and the projectile speed?

 

[IDumb]

meh, sorry, had some issues posting it must have gotten deleted.

I subbed in v_x = v₀cos($$\theta$$) and v_y = v₀sin($$\theta$$). Then i took dD/d$$\theta$$. Set it = 0, used a computer to solve it.

 

[NEILS BOHR]

at angle of 45

 

[gneill]

at angle of 45

Sure, if h = 0. What if h $$\neq$$ 0?

 

[IDumb]

Sure, if h = 0. What if h $$\neq$$ 0?

That's the issue. At h = 0 it simplifies easily. I get a messy equation for D with respect to theta. Leading to stupid mistakes when taking the derivative to find the critical points. I used a computer to solve what I had, and I got angles that varied with both height and the magnitude of the initial speed. That can't be right though, the angle shouldn't vary with the initial speed's magnitude.