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?

 

[tacman]

To find the angle for maximum x position, we can take the derivative of x1 with respect to [the] and set it equal to 0 to find the critical point. This will give us the angle at which the particle's x position is at a maximum.

Taking the derivative, we get:

dx1/d[the] = -v0*sin[the]*t1

Setting this equal to 0, we get:

-v0*sin[the]*t1 = 0

Since t1 is greater than 0, we can divide both sides by t1 to get:

-v0*sin[the] = 0

Solving for [the], we get:

[the] = 0 or 180 degrees

However, we know that the angle cannot be 180 degrees as this would result in the particle moving in the opposite direction. So, the only possible angle for maximum x position is [the] = 0 degrees.

Therefore, the particle will have maximum x position when it is initially launched at an angle of 0 degrees (horizontal). This makes sense as the horizontal velocity component is constant and does not change due to gravity, while the vertical velocity component decreases over time due to the acceleration of gravity.

In the case of y0 = 0, the problem is indeed easier to solve as the initial height does not affect the x position. In this case, any angle [the] would result in the same maximum x position.

I hope this helps!