How Does Train Acceleration Affect Projectile Firing Angle for Maximum Distance?

[Lamoid]

A cannon on a train car fires a projectile to the right with speed v relative to the train, from a barrel elevated at angle theta. The cannon fires just as the train, which had been cruising to the right along a level track with speed vt, begins to accelerate with acceleration a. Find an expression for the angle at which the projectile should be fired so that it lands as far as possible from the cannon. You can ignore the height of the cannon above the track.

Ok, so I know in the x direction

the projectile has a constant velocity of vcostheta + vt

in the y direction

initial veloctity of vsintheta.

I've tried solving for t in the vertical direction and then substituting into the x direction kinematic formula but I am still not sure how to find theta for some maximum distance.

Also, the answer has a (the accleration of the train) in it. I do not get at all why this is important unless it is talking about the case in which the a is so large that the train passes the projectile before it lands.

Thanks in advance.

 

[tiny-tim]

A cannon on a train car fires a projectile to the right with speed v relative to the train, from a barrel elevated at angle theta. The cannon fires just as the train, which had been cruising to the right along a level track with speed vt, begins to accelerate with acceleration a.
Find an expression for the angle at which the projectile should be fired so that it lands as far as possible from the cannon.

Hi Lamoid!

Hint: assume that the initial speed of the train is zero (in other words, use a frame of reference at the same speed as the train).

Then caclulate the position of the train and of the projectile at time t.

 

[thomate1]

I would approach this problem by first breaking it down into its components and considering the principles of projectile motion. We know that the horizontal motion of the projectile is independent of its vertical motion, so we can focus on finding the angle at which the projectile should be fired to achieve maximum horizontal distance.

First, let's consider the horizontal motion. As the train begins to accelerate, the projectile will also have a horizontal velocity of vcosθ + vt. This means that the horizontal displacement of the projectile can be represented by the equation x = (vcosθ + vt)t, where t is the time the projectile is in the air.

Next, let's consider the vertical motion. The initial vertical velocity of the projectile is vsinθ, and it will experience a constant downward acceleration of 9.8 m/s^2 due to gravity. We can use the equation y = vsinθt - 4.9t^2 to represent the vertical displacement of the projectile.

To find the angle at which the projectile should be fired, we need to find the time t at which the projectile reaches its maximum horizontal displacement. This occurs when the vertical displacement is at its maximum, which happens when the projectile reaches its highest point. Using the equation for vertical displacement, we can find that the time t at which this occurs is t = vsinθ/9.8.

Substituting this value of t into our equation for horizontal displacement, we get x = (vcosθ + vt)(vsinθ/9.8). To maximize this distance, we can take the derivative with respect to θ and set it equal to 0, giving us the equation vcosθ - vtsinθ = 0. Solving for θ, we get θ = tan^-1(vt/v).

Therefore, to achieve maximum horizontal distance, the projectile should be fired at an angle of tan^-1(vt/v). This angle will be affected by the acceleration of the train, as it changes the horizontal velocity of the projectile. If the acceleration is large enough, the train may pass the projectile before it lands, so it is important to consider the acceleration in this problem.

In summary, to find the angle at which the projectile should be fired for maximum horizontal distance, we use the equations for horizontal and vertical displacement and set the derivative of the horizontal displacement equal to 0. This angle will be affected by the acceleration of the train, so it