Simple trajectory, maximizing downhill range

In summary, a projectile fired down an inclined plane with an angle b with the horizontal and an initial speed of V can have its downhill range maximized at an angle of elevation a that is halfway between the plane and the vertical. This can be proven by finding the downhill range equation and using the coordinates of the intersection of the projectile's trajectory and the surface of the ground. Maximizing the downhill range also maximizes the x coordinate of the intersection.
  • #1
trajectory28
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Homework Statement


A projectile is fired from the origin down an inclined plane that makes an angle, b, with the horizontal. The angle of elevation of the gun and the initial speed of the projectile are a and V, respectively.

Show that the angle of elevation, a, that will mazimize the downhill range is the angle halfway between the plane and the vertical.




Homework Equations


I've already found the standard parametric equations of the trajectory as a function of time, t, below:

x(t)=(Vcos(a))t and y(t)=(Vsin(a))t - .5gt^2


The Attempt at a Solution


I've already found the downhill range to be:

[(Vcos(a))t]/cos(b)

and if the angle that maximizes the downhill range is indeed the angle halfway between the plane and the vertical, then:

(b+(pi/2))/2 = a + b

a = pi/4 - b/2

I just have no idea where to go after this, I've been trying to figure out a way to represent the time, t, at which the particle strikes the downhill plane as a function of V, a, and b. All has been to no avail so your guidance is much appreciated!
 
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  • #2
You can eliminate the time parameter from your first two equations to find y vs x instead of x(t) and y(t). You can also write an equation for y vs x representing the surface of the ground. The downhill range can be calculated from the coordinates of the intersection of these two expressions for y vs x. How does he downhill range relate to the x coordinate of the intersection. Does maximizing one of these maximize the other?
 
  • #3
Thank you!
 

1. What is a simple trajectory?

A simple trajectory is the path that an object takes when it is launched or thrown with a certain initial velocity and angle. It is a simplified representation of the object's motion, assuming no external forces act on it besides gravity.

2. How can I maximize downhill range?

To maximize downhill range, you need to find the optimal launch angle at which the object will travel the furthest distance on a downhill slope. This can be calculated using mathematical equations or through experimentation.

3. What factors affect the downhill range of an object?

The factors that affect the downhill range of an object include its initial velocity, launch angle, air resistance, and the slope of the terrain. These factors can be manipulated to maximize the object's distance traveled.

4. Can the downhill range be greater than the horizontal range?

Yes, it is possible for the downhill range to be greater than the horizontal range of an object. This occurs when the launch angle is less than 45 degrees, as the object will have a longer flight time and therefore travel a greater distance.

5. How is the downhill range related to the object's initial velocity?

The downhill range is directly proportional to the square of the initial velocity. This means that as the initial velocity increases, the downhill range will also increase. However, other factors such as air resistance and launch angle also play a role in determining the downhill range.

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