Q.A particle is projected into the air at some angle to the horizontal and moves along a parabolic trajectory, with $x$ and $y$ denoting the horizontal and vertical directions. Three points are marked on the path: A is on the rising part of the trajectory (early in the flight, still going up), B is at the very top of the trajectory (the highest point), and C is on the falling part (after the top, coming down). Describe the direction of the velocity and of the acceleration of the particle at each of the three points A, B and C.

Concept

Once airborne (neglecting air resistance), the only force on the particle is gravity, so its acceleration is constant: $\vec{a} = -g\,\hat{j}$ (straight down), regardless of where the particle is on its path. The velocity, in contrast, is tangent to the trajectory and changes direction continuously because the vertical component is being reduced (going up) or increased downward (coming down), while the horizontal component stays constant.

Velocity at each point

• At A (rising): the vertical velocity is upward and the horizontal velocity is forward, so $\vec{v}_A$ points up and forward, tangent to the curve, at an angle above the horizontal.
• At B (top): the vertical velocity is momentarily zero and only the (constant) horizontal component remains, so $\vec{v}_B$ is horizontal, pointing forward. …

Concept: One Continuously-Decreasing Angle Function Describes Velocity at Every Point on the Path

Method: A Single Monotonic Function $\theta_v(t) = \tan^{-1}\!\big(v_y(t)/v_x\big)$, Evaluated at Three Instants Instead of Argued Separately at A, B, C

Rather than describing the velocity direction at A, B, and C as three unrelated physical observations, this method writes the velocity's angle above the horizontal as one explicit, continuously changing function of time and shows it decreases monotonically throughout the flight -- so A, B, C are simply three points sampled along a single known curve, in a fixed order.

Step 1 -- Write $v_x$ and $v_y$ explicitly

With $u$ the launch speed and $\alpha$ the launch angle, and taking $t=0$ at launch:

Step 2 -- Define the velocity's angle above the horizontal as a single function of $t$

Since $v_x>0$ is fixed and $v_y(t)$ decreases linearly (from positive, through zero, to negative) as $t$ increases, and $\tan^{-1}$ is a strictly increasing function of its argument, $\theta_v(t)$ is a strictly decreasing function of time throughout the entire flight -- from a positive angle at launch, down through $0^\circ$, to a negative angle at landing. This one monotonic fact is all that's needed for every point on the trajectory, not just A, B, C.

Step 3 -- Locate A, B, C on this single decreasing curve

• At A (early, still rising): $t$ is small, so $v_y(t)=u\sin\alpha-gt$ is still positive $\Rightarrow \theta_v(t_A) > 0^\circ$. Velocity points up and forward.
• At B (the peak): by definition of the highest point, $v_y=0$ there $\Rightarrow \theta_v(t_B)=\tan^{-1}(0)=0^\circ$ exactly. Velocity is purely horizontal.
• At C (falling): $t$ has increased past $t_B$, so $v_y(t)=u\sin\alpha-gt$ has now gone negative $\Rightarrow \theta_v(t_C)<0^\circ$. Velocity points down and forward.

Because $\theta_v(t)$ was shown to be strictly decreasing in Step 2, these three signs are guaranteed to appear in exactly this order as $t$ increases -- there's no need to re-examine each point separately; they are simply three samples ($t_A<t_B<t_C$) of one already-known decreasing function.

Step 4 -- Acceleration needs no function at all -- it's a constant vector by Newton's second law

Once airborne, the only force is gravity, $\vec F=-mg\hat j$, so by $\vec F=m\vec a$: …