Kinematics question: At the maximum height, what velocity or speed is zero?

[LSD-25]

The Berkeley review's book is really confusing me. I know that the velocity is the X-direction is constant through an object's flight so that wouldn't be zero, so therefore the speed wouldn't be zero (speed is the magnitude of velocity vectors |v|=sqrt(vy^2+vx^2), but Vy (velocity is y direction) would be as that is the point where an object is at its maximum potential energy (KE=0 so V=0). Is all of that accurate? I thought it was but then I saw this question:
Consider a ball that is thrown straight up with a speed of v0. What are the magnitudes of its acceleration and speed when it reaches its maximum altitude?
A) a=0, v=v0
B) a=0, v=0
C) a=g, v=v0
D) a=g, v=0

The right answer is D, but I thought it was C. is it D because there's only a vertical component to the velocity, which becomes zero at its max height?

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[lifetothefullest]

There only is a vertical component of velocity at all heights. That's because it's thrown straight up. In other words, how would you ever get a horizontal velocity from throwing something straight into the air? The acceleration of some free object moving vertically is always gravity, so a = g is right and at the height, velocity = 0 because vertical velocity is zero and there is no horizontal velocity so everything is zero.

 

[LSD-25]

There only is a vertical component of velocity at all heights. That's because it's thrown straight up. In other words, how would you ever get a horizontal velocity from throwing something straight into the air? The acceleration of some free object moving vertically is always gravity, so a = g is right and at the height, velocity = 0 because vertical velocity is zero and there is no horizontal velocity so everything is zero.

good to know. if there was a horizontal component (e.g. thrown right 35m/s) the speed at the peak would be zero, vx would be 35m/s and vy=0 at its peak, right?

 

[lifetothefullest]

Well, like you said, the speed is just the magnitude of the velocity vector at any point, = sqrt(vx^2 + vy^2). So if the ball was thrown right with some initial vy and a vx of 35 m/s, at its peak, vx would be 35 m/s (assuming no air resistance) and vy would be 0 m/s but speed would be sqrt(vx^2) = vx = 35 m/s.

 

[ElectricNoogie]

For all projectile motion problems, unless the MCAT informs you about friction, you are free to ignore its effects. This means that the acceleration in the y directions is always g, and the acceleration in the x direction is always 0. Also, the Vy at the peak height will always be 0 m/s.