Question

A student drops a ball from a height of 72.0m. If the ball increases speed at a

uniform rate of 9.81m/s2, determine all unknowns and answer the following
questions.
How long did the ball remain in the air?

Answer 1

The ball remains in the air for approximately 3.02 seconds.

Step-by-step explanation:

To determine how long the ball remains in the air, we can use the kinematic equation:

d = vit + 1/2at^2

Where:

d = distance (72.0m)

vi = initial velocity (0 m/s)

a = acceleration (9.81 m/s^2)

t = time

Plugging in the values, we can rearrange the equation to solve for time:

72.0 = 0*t + 1/2*9.81*t^2

Simplifying and rearranging, we get:

4.905t^2 = 72.0

Solving for t, we find:

t^2 = 72.0/4.905

t = √(72.0/4.905)

t = 3.02 seconds

Therefore, the ball remains in the air for approximately 3.02 seconds.