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I'm doing the following problem:

A castle on a cliff has a cannon 300m above sea level. The cannon can shoot a 10kg iron ball with a velocity of 400 m/s . If the cannon is raised to an angle 30º of above the horizon, calculate the following:

the final velocity of the cannon ball just before it strikes the water

I know final Vy will be negative since it points down but doing the maths Vfinal is positive (and I think should be negative because it points down) and θ is negative. Is it correct? if yes. Why is θ negative?

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  • $\begingroup$ $v_f$ should be a positive nunber calculated using the Pythagorean theorem from $v_x$ and $v_y$ components $\endgroup$ – Triatticus Jan 30 at 0:09
  • $\begingroup$ typo corrected. It was positive on maths but my cuestion is why if its pointing down $\endgroup$ – Zaico Jan 30 at 0:14
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    $\begingroup$ It is a magnitude of a vector, it is strictly positive, the orientation is taken care of by theta, since the velocity should point down due to acceleration this is taken into account by theta $\endgroup$ – Triatticus Jan 30 at 0:29
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Velocity is a vector, and vectors have components. To find the magnitude of $V_{final}$ , we have to take the square root of both components squared.

$V_{final} = \sqrt{V_{x}^2+V_{y}^2}$

Because $V_{y}$ is squared, its sign doesn't really matter to the final answer. Also, the velocity cannot be negative, or else you would have to have imaginary velocities.

But because velocity is a vector, it also a direction. Here we use the formula $\theta = tan^{-1}(\frac {V_{y}}{V_{x}})$ to get the angle. The angle will be negative because the angle is from the x-axis. You have figured out that the projectile should be pointing downwards, but it is the angle that indicates that.

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Interestingly, the angle of the cannon above the horizon doesn't matter here at all, assuming no losses due to air friction. The ball exits the cannon with a specific kinetic energy. When the ball hits the water, the potential energy from the 300m height difference gets converted into additional kinetic energy.

If you are familiar with kinetic energy calculations, it is much easier to solve the problem by simply calculating the initial kinetic energy, then calculating the additional energy gained from the height difference, then converting the final kinetic energy back into the final velocity. No need to deal with angles and velocity components.

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  • $\begingroup$ Conservation of energy will give you the speed but not the direction so if the question requires velocity that would be incomplete. $\endgroup$ – M. Enns Jan 30 at 0:54
  • $\begingroup$ Oops, you are right here. You can still get the velocity vector this way (because you know both the - unchanged - horizontal speed and the total speed at splashdown), but it's probably no longer easier than calculating speed components directly. $\endgroup$ – cuckoo Jan 30 at 1:07
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Vfinal is positive because you're dealing with the module of the vector, module is always positive for any vector as is the sum of the squares of it's components, θ is negative because of the direction of the vector, if you draw the vector at the moment it touches the ground you'll see how θ is the angle between the x axis and the vector, which is negative because it's clockwise, while positive angles are anticlockwise.

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