A cannon ball whose barrel is angled at 40˚ from the horizontal fires a shell with a muzzle velocity of 350 m/s. What is the resultant velocity of the shell when it lands?

same as it was when it was shot.

To find the resultant velocity of the shell when it lands, we need to break the initial velocity of the shell into its horizontal and vertical components.

The horizontal component of the velocity remains constant throughout the projectile's motion. It can be determined using the equation:

Vx = V * cosθ

where Vx is the horizontal component of the velocity, V is the initial velocity of the shell, and θ is the angle of the cannon barrel from the horizontal.

Plugging in the values given in the question:

Vx = 350 m/s * cos(40˚) ≈ 267.87 m/s

The vertical component of the velocity changes due to the effect of gravity. It can be determined using the equation:

Vy = V * sinθ

where Vy is the vertical component of the velocity.

Plugging in the values given in the question:

Vy = 350 m/s * sin(40˚) ≈ 224.85 m/s

Now, we can find the time taken to reach the highest point of the projectile's trajectory using the equation:

t = Vy / g

where g is the acceleration due to gravity (approximately 9.8 m/s²).

Plugging in the values:

t = 224.85 m/s / 9.8 m/s² ≈ 22.94 s

Since the shell takes the same amount of time to reach the highest point and descend, the total time of flight is 2 * t ≈ 45.88 s.

Finally, to find the resultant velocity when the shell lands, we need to find the vertical component of the velocity at that time. We can use the equation:

V = Vy - g * t

Plugging in the values:

V = 224.85 m/s - 9.8 m/s² * 45.88 s ≈ -223.12 m/s (negative sign indicates downward direction)

Therefore, the resultant velocity of the shell when it lands is approximately 223.12 m/s in the downward direction.