The range of a projectile fired up the side of a mountain?

[cj]

Another question pertaining to a lab that I have to setup...

The range of a projectile fired up the side of a mountain?

At the base of a mountain with a constant slope of
$$\phi$$, a projectile is fired from a cannon oriented
at an angle of $$\theta$$.

What $$\theta$$ will result in the maximum range up the
side of the mountain?

What is this maximum range??

- air resistance can be neglected.

---
As a starting point I'm looking at

$$x_{max}=\frac{v_0^2sin2\alpha}{g}$$

But am not clear about next steps.

 

[NateTG]

This is a classic problem.

That said, it's not at all clear that your formula applies to this situation. Perhaps it's easier to calculate the distance the projectile travels in terms of $$\phi$$,$$\theta$$ and $$v_0$$ and then find the maximum?

 

[cj]

Yes, this is the line of thinking I was considering -- but came to a dead-end.

This is a classic problem.

That said, it's not at all clear that your formula applies to this situation. Perhaps it's easier to calculate the distance the projectile travels in terms of $$\phi$$,$$\theta$$ and $$v_0$$ and then find the maximum?

 

[arildno]

Find the equation for the parabola with the correct slope (y as a function of x).
Find the intersection of this parabola with the straight line representing the mountain slope (i.e, find the x-coordinate of the point of intersection).

alternatively, write the equations of motion in terms of normal and tangential components to the mountainside. Determine the time at which the projectile's normal position is again 0.

Anyhow, once you've found this, you may find an equation for distance as a function of the angle that can be maximized.

 

[kurious]

Form a triangle so that you have the slope representing the distance traveled up the mountain.
(u cos theta) t is horizontal distance (theta = angle between initial velocity vector,u, and horizontal) and 1/2 gt ^ 2 is vertical distance on triangle.

distance up mountain = sqrt [ (1/2gt ^2)^ 2 + (u cos theta)^2 t ^ 2]

Since distance up mountain is a function of time:
for maximum distance up mountain

d(distance)/ dt = 0

so d/dt sqrt [ (1/2gt ^2)^ 2 + (u cos theta)^2 t ^ 2]= 0

If phi is angle of mountain slope, then
tan phi = vertical of triangle/ horizontal of triangle
tan phi = 1/2 gt ^ 2 / ( u cos theta) t
so 1/2 gt^2 = (u cos theta) t x tan phi
u cos theta = (1/2 gt)/tan phi

so d/dt sqrt [ (1/2gt ^2)^ 2 + (1/2 gt) ^2/ (tan phi)^2 t ^ 2]= 0
d/dt sqrt [ 1/4 g^2 x t^4 + 1/4g^2 x t^2/ (tan phi)^2 t^2 ]= 0

for constant phi we can evaluate d/dt.


( differentiation can be done using u substitution and chain rule)

substituting the theta and phi for t in the differentiated expression
will give maximum distance projectile travels up mountain.