Question

An object falls from a position of rest and reaches an instantaneous speed of 7 m/s. How far will it have dropped? The acceleration of gravity is 9.8 m/s. What formula should be used to solve this problem?

A) Distance = (Initial Velocity * Time) + (0.5 * Acceleration * Time^2)
B) Distance = (Final Velocity * Time) + (0.5 * Acceleration * Time^2)
C) Distance = (Initial Velocity * Time) + (Acceleration * Time^2)
D) Distance = (0.5 * Acceleration * Time^2)

Answer 1

The correct formula to use for an object dropped from rest is Distance = (0.5 * Acceleration * Time^2). Solving for time using the final velocity and acceleration due to gravity, and then using that time to calculate the distance, gives a drop distance of approximately 2.47 meters.

Step-by-step explanation:

The appropriate formula to determine how far an object has dropped when it falls from a position of rest and reaches an instantaneous speed of 7 m/s, under the acceleration of gravity, which is 9.8 m/s2, is as follows:

D) Distance = (0.5 * Acceleration * Time2)

This is because the initial velocity (vo) of the object is zero since it falls from a position of rest, meaning the typical term (Initial Velocity * Time) does not contribute to the distance traveled. Thus, we only consider the displacement due to acceleration over time.

To solve for the time taken to reach the speed of 7 m/s:

v = vo + at

7 m/s = 0 m/s + (9.8 m/s2 * t)

t = 7 m/s / 9.8 m/s2 = 0.714 seconds approximately

Plugging the time back into the distance formula:

D = 0.5 * 9.8 m/s2 * (0.714 s)2

D ≈ 2.47 meters

The object would have dropped approximately 2.47 meters.