# Projectile motion - surface of max range

• insynC
In summary, in an Analytical Mechanics course, the question of a gun firing shells at any direction with the same speed v and reaching a surface of z = v²/(2g) - ρ²g/(2v²) was discussed. Using conservation of energy, the maximum height of the bullet at a given angle θ was found to be z = (vsinθ)²/(2g) and the range was found to be ρ(max z) = 2v²/g sinθcosθ. However, this approach was found to be complicated and a suggestion was made to start again without taking any shortcuts.
insynC
Doing an Analytical Mechanics course and we're just reviewing some classical physics and this question came up. Cheers for any help.

## Homework Statement

A gun can fire shells in any direction with the same speed v. Ignoring air resistance and using cylindrical polar coordinates with the gun at the origin and z measured vertically up, show that the gun can hit any object inside the surface:

z = v²/(2g) - ρ²g/(2v²)

(where ρ & z are as per cylindrical coordinates)

## The Attempt at a Solution

My thought was that the surface is the set of points which are the farthest the gun can reach. As such the max vertical height can be found by conservation of energy:

Max height when all vertical component of KE turned into PE.

i.e. if θ is the angle trajectory from the ground:

1/2 m (vsinθ)² = mgz

Rearranging:

z = (vsinθ)²/(2g) , which I think gives the max height the bullet will reach for a given θ. (Note that if θ = 90° => ρ = 0 gives z = v²/(2g) as per the answer).

Next I wanted to determine ρ at the max height. As the motion will be parabolic, the max height should be achieved at half of the range for a given θ.

Invoking Newton's second law and bashing through algebra an equation for z in terms of ρ can be found (by eliminating θ), and then setting z = 0 a solution for ρ(range) can be found, which when halved is ρ(max z):

ρ(max z) = 2v²/g sinθcosθ

Where I'm at

At this stage I have tried bashing these two things around a bit with little success in deriving the given result. I feel the problem shouldn't be this complicated and that I may have taken a wrong turn early on.

Any suggestions would be greatly appreciated.

Hi insynC!
insynC said:
Next I wanted to determine ρ at the max height. As the motion will be parabolic, the max height should be achieved at half of the range for a given θ.

I feel the problem shouldn't be this complicated and that I may have taken a wrong turn early on.

Yes … you're finding the range at ground level

start again, and no short-cuts, this time!

Thanks for the help, it came out.

## 1. What is projectile motion?

Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity and air resistance.

## 2. How does the surface affect the maximum range of a projectile?

The surface does not have a significant effect on the maximum range of a projectile. The maximum range is determined by the initial velocity, launch angle, and acceleration due to gravity.

## 3. What is the optimal launch angle for maximum range?

The optimal launch angle for maximum range is 45 degrees. This angle allows for the maximum horizontal displacement while still maintaining a significant vertical displacement.

## 4. How does air resistance affect projectile motion?

Air resistance can reduce the maximum range of a projectile by slowing it down and altering its trajectory. It can also cause the projectile to deviate from its expected path.

## 5. Can a projectile have a negative range?

Yes, a projectile can have a negative range if it is launched at a downward angle. This means that the projectile will land behind the initial launch point.

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