Recoiling Cannon Projectile Velocity

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SUMMARY

The discussion focuses on the physics of a circus cannon firing a projectile at an initial velocity of 40 m/s while recoiling at 1.5 m/s. The cannon has a mass of 4000 kg and is tilted at an angle of 45°. To determine the angle of the projectile's motion with respect to the ground, the horizontal and vertical components of the projectile's velocity must be resolved. The solution involves applying the principle of conservation of momentum and vector resolution to find the resultant velocity and angle.

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Homework Statement



A circus cannon, which has a mass M = 4000 kg, is tilted at q = 45°. When it shoots a projectile at v0 = 40 m/s with respect to the cannon, the cannon recoils along a horizontal track at vcannon = 1.5 m/s with respect to the ground.

a) At what angle to the horizontal does the projectile move with respect to the ground?
(The angle is NOT 45°)

b) What is the mass of the projectile?

c) The cannon is now lowered to shoot horizontally. It fires the same projectile at the same speed relative to the cannon. With what speed does the cannon now recoil with respect to the ground?

Homework Equations


p = mv

The Attempt at a Solution



I have no idea where to begin to solve for the angle.
 
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pleasehelpme6 said:

Homework Statement



A circus cannon, which has a mass M = 4000 kg, is tilted at q = 45°. When it shoots a projectile at v0 = 40 m/s with respect to the cannon, the cannon recoils along a horizontal track at vcannon = 1.5 m/s with respect to the ground.
.
When the projectile leaves the cannon, it has two velocities. Vo at angle 45o and Vcannon in the horizontal direction.

Resolve Vo into two components, Vocos(45) in the horizontal direction and Vosin(45) in the vertical direction.

Now find the net horizontal velocity. Then the resultant velocity and the angle of the resultant velocity with the horizontal.
 

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