Calculating the Angle of Projection for a Fired Cannon Shell

[rukawa1107]

A 1250 kg cannon, which fires a 55 kg shell with a speed of 566 m/s relative to the muzzle, is set at an elevation angle of 39° above the horizontal. The cannon is mounted on frictionless rails, so that it recoils freely.

What is the speed of the shell with respect to the Earth?

Vse=Vsm+Vme, Vme=Vse-Vsm, Vsm= 566m/s

MVi= 0 = 55kg(Vse)+1250kg(Vse-566m/s)

so, Vse=542m/s and Vme = 542m/s-566m/s = -24m/s

I have solved the first part by using the conservation of momentum.

However, I have some diffculty on the second part.


At what angle with the ground is the shell projected?

This angle of the shell relative to the ground can be given by the ratio

calculation however I am not so sure.

Can anyone please solve and show me how to solve the second part?

 

[Doc Al]

Careful: Momentum is conserved in the horizontal direction only.

Once you correctly find the recoil speed of the cannon, you can find the velocity of the ball with respect to the Earth via:
[tex]\vec{V}_{b/e} = \vec{V}_{b/c} + \vec{V}_{c/e} [/tex]

(It seems that Latex is not displaying. To find the velocity of the ball with respect to the Earth find the vector sum of the velocity of the cannon wrt Earth plus the velocity of the ball wrt the cannon.)

 

[kyle8921]

To find the angle of projection of the shell, we can use the equation for projectile motion:

θ = tan⁻¹(Vy/Vx)

Where θ is the angle of projection, Vy is the vertical component of the velocity, and Vx is the horizontal component of the velocity.

In this case, we know the initial velocity of the shell (Vse = 542m/s) and the elevation angle (39°). We can use trigonometric functions to find the vertical and horizontal components of the velocity:

Vy = Vse * sin(39°) = 542m/s * 0.6293 = 341.7m/s
Vx = Vse * cos(39°) = 542m/s * 0.7771 = 421.1m/s

Now, we can substitute these values into the equation for θ:

θ = tan⁻¹(341.7m/s / 421.1m/s) = tan⁻¹(0.811) = 39.6°

Therefore, the angle of projection of the shell with respect to the ground is 39.6°. We can also check this by using the conservation of energy: the kinetic energy of the shell at launch should equal the potential energy at the highest point of its trajectory. Using the equation for potential energy, we can calculate the maximum height of the shell:

PE = mgh
h = PE / mg = (55kg * 9.8m/s² * 421.1m/s) / (55kg * 9.8m/s²) = 421.1m

This means that the shell will reach a maximum height of 421.1m, and at this point, its velocity will be purely vertical. Therefore, the angle of projection with respect to the ground must be 90° - 39° = 51°. This is close to the value we calculated earlier, which confirms our answer.