Velocity at the top of the path

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SUMMARY

In projectile motion, when a projectile reaches its maximum height, its vertical velocity (v_y) is zero while its horizontal velocity (v_x) remains constant, assuming no horizontal acceleration (a(x) = 0). The overall velocity at this point is not zero, as it is equal to the horizontal velocity (|v| = |v_x|). The only scenario where the overall velocity is zero occurs when both the horizontal and vertical velocities are zero, such as when a projectile is dropped straight down or shot vertically without any horizontal component.

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OK, simple question, but I need clarification. When a projectile reaches its maximum height on its trajectory, it's vertical velocity is 0, but it's horizontal velocity is not, right? Because the horizontal velocity of a projectile (assuming a(x)=0, and a(y)=-g) is constant. Thus, the overall velocity at that moment is not zero, right? Thanks!
 
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That's correct. Usually in projectile motion, the horizontal velocity remains constant throughout the motion (unless, for example, you take friction into account). In the special case where this constant is zero, the projectile goes straight up and down (for example, a ball bouncing from the floor or a bullet being shot straight up). At the top of the motion, the overall velocity
|v| = \sqrt{v_x^2 + v_y^2} \stackrel{v_y = 0}{=} |v_x|
is at a minimum. Also here you can see, that |v| = 0 if and only if the horizontal velocity vx = 0.
 

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