Projectile Speed Comparison: Explaining with No Equations

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When two objects are fired from the same height and initial speed, their final speeds upon landing will be the same, despite differing angles of projection. The discussion emphasizes that energy conservation principles apply, meaning potential and kinetic energy at launch and landing remain equal. Although the velocity vectors differ in their horizontal and vertical components, the overall speed at impact is unaffected by the launch angle. The key takeaway is that both objects, regardless of their angles, will land with the same speed. This conclusion can be reached without resorting to equations, relying instead on the concept of energy conservation.
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If two objects are fired, both from a height h and speed v0, where object A is fired at an angle θ with the horizontal and object B is fired at an angle of 2θ, which will land with a greater speed?

It seems to me that the speed would be the same for both, although the velocity vectors would differ in their x- and y-components. Is my answer correct? The question asks to explain this without any equations, so I am having trouble justifying it. Any ideas? Thank you.
 
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HINT: Energy is conserved! :)
 
So if I use y=0 as my reference point and y=h as the launching point, I can say that:

Object A:

mgh+\frac{1}{2}m{v_{0}}^{2}=\frac{1}{2}m{v_{A}}^{2}

Object B:

mgh+\frac{1}{2}m{v_{0}}^{2}=\frac{1}{2}m{v_{B}}^{2}

Then vA must equal vB. Is this correct?
 
Yes, and you can even say that without using equations! :)
 
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