Why is acceleration still negative at the top of a throw

Constant Gravity at the Apex

When an object is thrown straight up, the only force acting after it leaves the hand is gravity. Gravity does not depend on the object's speed or direction; it always points toward the Earth with magnitude $g \approx 9.8\ \text{m/s}^2$. If we choose upward as the positive direction, the acceleration vector is $-g$, which is negative. At the highest point the velocity momentarily becomes zero, but the acceleration stays $-9.8\ \text{m/s}^2$ because the gravitational force has not vanished.

Common misconceptions about the top of a throw

• Zero velocity means zero acceleration.
• Gravity disappears at the highest point.
• A negative sign only indicates slowing down, not direction of force.

Finding the top of the trajectory for a thrown ball

1. 1Write the velocity equation: $v(t) = v_0 - g t$.
2. 2Set $v(t) = 0$ to locate the apex and solve for time: $t_{top} = \frac{v_0}{g}$.
3. 3Insert $t_{top}$ into the position equation $y(t) = y_0 + v_0 t - \frac{1}{2} g t^2$ to obtain the maximum height.

Sign convention versus actual values at the apex

A concrete example makes this clear. Throw a ball upward with $v_0 = 20\ \text{m/s}$. The time to reach the top is $t_{top}=20/9.8 \approx 2.04\ \text{s}$. At that instant the speed is 0, yet the acceleration remains $-9.8\ \text{m/s}^2$, pulling the ball back down. This constant negative acceleration is why the ball begins to descend immediately after the apex, and why the motion equations use the same $-g$ term throughout the entire flight.