Cannon fired from moving truck

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Discussion Overview

The discussion revolves around determining the optimal angle for firing a cannon from a moving truck to achieve the maximum horizontal range of the projectile. Participants explore different methods of calculation, including geometric and calculus-based approaches, while addressing assumptions about the motion of the truck and the cannon.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant claims that using a geometric method involving triangles yields an optimal angle of 49 degrees for maximum range.
  • Another participant argues that a calculus-based approach results in an angle of 46.9 degrees, suggesting that this method accounts for the additional horizontal velocity provided by the moving truck.
  • Some participants express confusion about the triangle method, questioning its validity given that projectile motion does not conform to a simple triangular model.
  • There is a claim that the maximum range for projectiles is typically at 45 degrees, but this may not apply when the cannon is fired from a moving vehicle, as the initial horizontal velocity alters the optimal angle.
  • One participant suggests that the effective angle for maximum range should be recalculated, proposing an angle of approximately 45.17 degrees based on their interpretation of the problem.
  • Another participant emphasizes that the velocity of the projectile should be treated as a vector, combining the truck's speed with the cannon's firing angle to determine the total velocity components.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the optimal angle for maximum range, with multiple competing views and methods presented. The discussion remains unresolved regarding which approach is correct or if a definitive answer exists.

Contextual Notes

Participants note that the differing results may stem from varying assumptions about the problem setup, particularly regarding the initial velocities and the interpretation of the projectile's motion. The discussion highlights the complexity of projectile motion when influenced by an additional horizontal velocity.

esmikell
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Hello everyone,

I have a physics question for you all. A cannon fired from a truck at 5m/s. A cannon is fired in the same direction as the cannon. The velocity of the projectile is 50m/s. What angle will provide the greatest (distance in the x-driection) range for the projectile.

I have gotten two solution.
Using triangles and that both have the same maximum height the best angle is 49 degrees.

Using calculus we have come with a best angle of 46.9 degrees.

Which is best and why? I believe it is calculus because the calculus version is assuming the cannon fired at the non-45 degrees will have a higher altitude. Therefore ruining the triangles method.

Thank you.
 
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I have no idea what you mean by using "triangles". Clearly at least one of these methods must be wrong if they get different answers, so they must be based on different assumptions. Describe both solutions.
 
The triangles method uses a 45-45-90. From the top angle drop a line down to the base leg. The left side is 5 and the other is x. The side in common is h, the maximum height. Since this is a 45-45-90, the 5+x side is equal to the h. There is enough variables equal to each other that the angle in the x-h triangle can be found to be 49 degrees.

However, change it to using the x=vt equation and things change. Since we are giving the projectile the extra x-velocity I wrote the velocity in the x-direction to be:

vx= vtruck+vcos(theta) v=the velocity of the projectile

x=(vtruck+vcos(theta)t

I want to do a derivative on theta and find the zero to get the max distance. So I need to get rid of t.

t is found using vf=vi+at. All of this in the y-direction

t=2vsin(theta)/g again v is projectile velocity

So after some substitutions, cleaning up, taking a derivative and getting a sin^2 and a cos^2, I found an equation with thetas hard to pull out.

I put the equation into excel and forced angles into the equation until my derivative equation hit zero, therefore, showing me the angle that creates the maximum distance.

This method gives 46.9 as theta.

I hope this clears up what I was asking. Thank you for your time.

Stephen
 
esmikell said:
The triangles method uses a 45-45-90. From the top angle drop a line down to the base leg. The left side is 5 and the other is x. The side in common is h, the maximum height. Since this is a 45-45-90, the 5+x side is equal to the h. There is enough variables equal to each other that the angle in the x-h triangle can be found to be 49 degrees.
Since the trajectory of the cannonball is not a triangle, this method doesn't make any sense to me.

However, change it to using the x=vt equation and things change. Since we are giving the projectile the extra x-velocity I wrote the velocity in the x-direction to be:

vx= vtruck+vcos(theta) v=the velocity of the projectile

x=(vtruck+vcos(theta)t

I want to do a derivative on theta and find the zero to get the max distance. So I need to get rid of t.

t is found using vf=vi+at. All of this in the y-direction

t=2vsin(theta)/g again v is projectile velocity

So after some substitutions, cleaning up, taking a derivative and getting a sin^2 and a cos^2, I found an equation with thetas hard to pull out.

I put the equation into excel and forced angles into the equation until my derivative equation hit zero, therefore, showing me the angle that creates the maximum distance.

This method gives 46.9 as theta.
Looks good to me. Just for the challenge of it, try to solve for theta analytically. (Express that last equation completely in terms of cosines. You'll get a quadratic that you should be able to solve easily. You'll get the same answer of course.)
 
Good morning,

Thanks for looking into it. I will see what I can do with the cos's.

Stephen
 
First of all, maximum horizontal projectile range is always 45 degrees.

Second, the horizontal range is given by (v^2/g) Sin 2(theta)...


these can be derived from x (range) =vt cos(theta)

and y (vertical) = vtsin(theta) - 1/2 gt^2

and of course "t" time of flight is the same for both x and y...
 
Max range is always at 45 when you projectile is launched from an object that is not moving in the x-direction.

If you fire a cannon that is on a railcar moving with a velocity, a 45 degree cannon will not give a max range. Since there is already a x-velocity you can then steal some x-velocity from the cannon and give it to the y-velocity; therefore, giving it more time in air; therefore, more time to run in the x-direction.
 
Naty1 said:
First of all, maximum horizontal projectile range is always 45 degrees.

Second, the horizontal range is given by (v^2/g) Sin 2(theta)...
Only for a fixed speed (v), which is not the case here. (The speed of the cannonball with respect to the ground depends upon the angle.)
 
esmikell said:
Hello everyone,

I have a physics question for you all. A cannon fired from a truck at 5m/s. A cannon is fired in the same direction as the cannon. The velocity of the projectile is 50m/s. What angle will provide the greatest (distance in the x-driection) range for the projectile.
Can we assume that you mean the TRUCK is moving at 5 m/s and that the cannon is fired in the same direction as the truck is moving?

I don't get "49" or "46.9" degrees. I get approximately 45.17 degrees.
 
Last edited by a moderator:
  • #10
HallsofIvy said:
Can we assume that you mean the TRUCK is moving at 5 m/s and that the cannon is fired in the same direction as the truck is moving?
Yes. And that the cannonball speed is 50 m/s with respect to the truck.
I don't get "49" or "46.9" degrees. I get approximately 45.17 degrees.
Do it over. :wink:
 
  • #11
Hello HallsofIvy,

The cannon is on the truck. The truck is moving at 5m/s, let's say moving to the east. The cannon is fired with an unknown angle. The projectile has a velocity of 50 m/s at the unknown angle. The cannon is fired in the east direction.
 
  • #12
just add 5 ms/sec cos(theta) and 5m/s sine (theta) to the fifty m/s to determine total projectile speed...for altitude or range or whatever you want to play with...the projectile doesn't know if the cannon is moving or if it has a slightly bigger charge...
 
  • #13
The velocity vector for the projectile on a moving railcar, truck, whatever is going to be

v = [5+vcos(theta)]i + [vsin(theta)]j

The 5m/s is added to the i component (x-dir). You can't go and add 5m/s to the y-dir.
 

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