A particle is thrown vertically upward. When it reaches the highest point of its motion and is instantaneously at rest, what is its acceleration at that point?

Introduction / Context:
This question tests understanding of motion under gravity, specifically what happens at the highest point when an object is thrown vertically upward. Many learners know that the velocity becomes zero at the top, but they sometimes misunderstand the behaviour of acceleration at that instant. Clarifying this point is very important for correctly solving vertical motion problems in kinematics.

Given Data / Assumptions:

• A particle is thrown straight up in the vertical direction.
• We consider motion in a uniform gravitational field near the Earth.
• Air resistance is neglected.
• We are interested in the state of acceleration at the highest point.

Concept / Approach:
Under uniform gravity near the Earth surface, the acceleration due to gravity g acts downward at all points along the path, irrespective of the direction of motion. When the particle reaches the highest point, its instantaneous velocity becomes zero, but gravity is still acting. Therefore, the acceleration remains g downward. The fact that velocity is zero at that instant does not mean acceleration is zero. The presence of downward acceleration at the top causes the particle to start moving downward again.

Step-by-Step Solution:
Step 1: Recognise that throughout the motion, the only significant force on the particle is its weight, mg, acting downward. Step 2: This constant weight produces a constant acceleration g downward, regardless of where the particle is on its path. Step 3: At the highest point, the vertical component of velocity becomes zero, but the force mg is still present. Step 4: Since acceleration is net force divided by mass, and mg is non zero, the acceleration cannot be zero. Step 5: The direction of this acceleration is downward, towards the Earth, with magnitude approximately 9.8 m/s^2. Step 6: Therefore, the correct description is that the particle has a downward acceleration at the highest point. Step 7: As a result of this downward acceleration, the particle begins to move downward immediately after reaching the highest point.

Verification / Alternative check:
Kinematic equations such as v^2 = u^2 + 2 * a * s can be applied. From the start to the highest point, the acceleration a is constant and equal to g downward (negative if upward is taken as positive). At the top, v = 0, but we use the same value of a as in the rest of the motion. If acceleration were zero at the top, the particle would stay there forever once it stopped, which does not happen in reality. Instead, it immediately falls back, proving that acceleration remains downward g at the highest point.

Why Other Options Are Wrong:
An upward acceleration would require a net upward force at the highest point, which does not exist in simple free fall under gravity. A downward velocity is incorrect at the highest point, because the velocity at that instant is zero just before changing direction. A horizontal velocity is not part of the described purely vertical motion, so it does not apply here.

Common Pitfalls:
A frequent misconception is to think that whenever velocity is zero, acceleration must also be zero. This is not true in general. Velocity can be zero at an instant while acceleration is non zero; for example, at turning points in vertical motion or at the extremes of simple harmonic motion. To avoid this mistake, always think in terms of forces. If a net force acts, acceleration is present even if instantaneous velocity is zero. In vertical throws, gravity always acts downward, so acceleration is always downward g.

Final Answer:
At the highest point of its motion, the particle has a downward acceleration equal to g due to gravity.