Question

A ball is released from rest from an elevated position and free-falls towards the ground. Assuming negligible air resistance, what distance does the ball fall during the first 3.3 seconds?

Answer 1

Using the kinematic equation for uniformly accelerated motion (d = ½gt²), the ball falls approximately 53.37 meters during the first 3.3 seconds, assuming an acceleration due to gravity of 9.8 m/s² and negligible air resistance.

Step-by-step explanation:

To determine the distance a ball falls in free fall during the first 3.3 seconds, we can use the kinematic equation for uniformly accelerated motion without initial velocity:

d = ½gt²
Where d is the distance, g is the acceleration due to gravity (9.8 m/s²), and t is the time in seconds. Substituting the given values:

d = ½(9.8 m/s²)(3.3 s)²

Calculating the distance:

d = 0.5 * 9.8 m/s² * (3.3 s)² = 0.5 * 9.8 m/s² * 10.89 s² = 53.367 m

Therefore, the ball falls a distance of approximately 53.37 meters during the first 3.3 seconds.