Plotting Hyperbolic Curve from a Moving Train's Bullet

[snazzy1981]

A train is traveling at the speed of a bullet.

A man, stood on top of the train, fire's a gun in the direction from which the train has come from. (facing rearward)

He pulls the trigger the instant the train passes station 'A'.

The velocity of the bullet relative to the ground is 0

To an observer standing on the platform (mind you, in a perfect vacuum) the bullet would drop.

In our imperfect world, the bullet would drop about 2/3 of the way and then follow the train in a hyperbolic curve.

would someone mind plotting curve ?. I am a simple web developer and this has got me curious as it was a question posted on my own forum... thanks

 

[fatra2]

Hi there,

Why would your bullet, in our imperfect world, would travel 2/3 of the way. The bullet is fired backward at the speed of the train => the bullet's speed compared to someone in the station would be 0. I don't see any reason for it to suddenly be bothered by friction, since it has no velocity compared to the still air surrounding. Of course, I base my assumption on the fact that the bullet leaves the barrel at the speed of the train, friction in the barrel not considered.

Can you explain the reason for this. Cheers

 

[sir-pinski]

Sure, I'd be happy to help plot the hyperbolic curve for this scenario. First, let's break down the information we have:

- The train is traveling at the speed of a bullet, which is typically around 1,700 meters per second.
- The man on top of the train fires a bullet in the direction the train has come from, which we can assume is also the direction the train is currently traveling in.
- The man pulls the trigger the instant the train passes station A, which means the train is already in motion when the bullet is fired.
- The velocity of the bullet relative to the ground is 0, meaning it is not moving horizontally.
- The bullet will drop due to gravity, but because it is fired from a moving train, it will follow a hyperbolic curve instead of a parabolic curve.

To plot the hyperbolic curve, we will need to use the equation for a hyperbola, which is:

x^2/a^2 - y^2/b^2 = 1

Where a and b are the distances from the center to the vertices of the hyperbola. In this case, the center of the hyperbola will be the point where the bullet is fired (station A), and the vertices will be the points where the bullet hits the ground.

To find a and b, we can use the following equations:

a = (v^2/g) * sin(2θ)

b = (v^2/g) * cos^2(θ)

Where v is the initial velocity of the bullet (in this case, the speed of the train, 1,700 m/s), g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of elevation of the bullet when it is fired.

Since the bullet will drop 2/3 of the way and then follow the train, we can assume that the angle of elevation is 45 degrees (since the angle of elevation for a projectile at maximum range is 45 degrees).

Plugging in the values, we get:

a = (1700^2/9.8) * sin(90) = 291,428.57 meters

b = (1700^2/9.8) * cos^2(45) = 241,071.43 meters

Now, we can plot the hyperbolic curve on a graph, with the x-axis representing