Projectile motion concept: A particle is launched with speed v at an angle α above the horizontal. At which point in its flight is the centripetal (normal) component of acceleration maximum?

Introduction / Context:

In projectile motion under uniform gravity, total acceleration is constant and equals g downward. However, its decomposition into tangential (along velocity) and normal/centripetal (perpendicular to velocity) components varies along the path. Understanding this split is useful in dynamics and flight trajectories.

Given Data / Assumptions:

• No air resistance; acceleration due to gravity g is constant and vertical.
• Velocity direction changes along the parabolic path.

Concept / Approach:

Resolve g into components relative to the instantaneous velocity direction. Let ψ be the flight-path angle (angle of velocity with horizontal). Then tangential acceleration a_t = g * sin(ψ) and normal (centripetal) acceleration a_n = g * cos(ψ). Since |cos(ψ)| ≤ 1, the maximum possible a_n is g, attained when ψ = 0°, i.e., when the velocity is horizontal—at the top of the trajectory.

Step-by-Step Solution:

Verification / Alternative check:

An equivalent view uses curvature κ and speed v: a_n = v^2 * κ. At the top, although speed is minimum, the curvature is maximum for a projectile, and the product equals g—consistent with the component resolution result.

Why Other Options Are Wrong:

• At start/impact, ψ = ±α, so a_n = g * cos α < g.
• “Elsewhere” is incorrect because the maximum is specifically at the peak.
• “Constant” is wrong; only total acceleration is constant, not its components.

Common Pitfalls:

• Confusing maximum speed with maximum normal acceleration; they occur at different points.

Final Answer:

at the top of the trajectory