Acceleration of a stone thrown upwards

Problem

You throw a stone upwards with velocity $v_0$. A friend claims that, while the stone is rising, the acceleration points upwards because it “follows the motion”. Are you right or is he? Discuss by drawing the velocity and the acceleration at three instants: rising, top, falling.

Solution

You are right. Neglecting air, only gravity acts on the stone: the acceleration is $\vec{a} = \vec{g}$, constant, of magnitude $\approx 9{,}81\;\text{m/s}^2$ and always directed downwards, regardless of whether the stone is rising, momentarily at rest at the top, or falling.

$\ev{\vec{a} = \vec{g} \; \text{constant, always downwards}}$

The friend’s mistake. He confuses the direction of the velocity with that of the acceleration. The acceleration does not “follow the motion”: it indicates how the velocity changes. While rising the velocity points upwards but is decreasing, so its variation (and hence the acceleration) points downwards.

The three instants.

• Rising: $\vec{v}$ upwards and decreasing; $\vec{a}$ downwards (opposite to $\vec{v}$): the stone decelerates.
• Top: $\vec{v} = 0$ for an instant; $\vec{a}$ still downwards, $\approx 9{,}81\;\text{m/s}^2$. It is precisely this non-zero acceleration that makes the stone start moving downwards again.
• Falling: $\vec{v}$ downwards and increasing; $\vec{a}$ downwards (in the same direction as $\vec{v}$): the stone accelerates.

At all three instants the red acceleration arrow points downwards: only the velocity changes.

Collegamenti

Argomenti: Kinematics Concetti: Free fall · Acceleration