Question

t = 0 s, d = 0 m →
t = 1 s, d = 4.9 m →
t = 2 s, d = 19.6 m →
t = 3 s, d = 44.1 m →
t = 4 s, d = 78.4 m →
t = 5 s, d = 123 m →
q. what happens to the velocity of a ball as it is dropped off a cliff?

Explanation:

Step1: Recall free - fall motion

In free - fall, the acceleration of an object near the Earth's surface is approximately $g = 9.8m/s^{2}$ (neglecting air resistance). The velocity of an object in free - fall is given by the formula $v=v_{0}+gt$, where $v_{0}$ is the initial velocity, $g$ is the acceleration due to gravity, and $t$ is the time.

Step2: Analyze the initial condition

The ball is dropped, so $v_{0} = 0m/s$. Then the velocity formula simplifies to $v = gt$. As time $t$ increases (since the ball is falling for longer periods as it descends), and $g$ is a constant ($g=9.8m/s^{2}$), the velocity $v$ of the ball increases.

Answer:

The velocity of the ball increases as it is dropped off a cliff because it is in free - fall and the acceleration due to gravity causes its speed to increase over time.