An egg is thrown downward at [tex]$3.15 \, \text{m/s}$[/tex] from the roof of a [tex]27.5 \, \text{m}$[/tex] tall building.

What is the final velocity of the egg just before it hits the ground?

[tex]v_{f} = [\text{?}] \, \text{m/s}[/tex]

Do not account for air resistance. Remember, velocity downward is a negative vector [tex](-)[/tex].



Answer :

Certainly! To determine the final velocity of an egg thrown downward from a roof, we can use the kinematic equations of motion. Let's break the problem into clear steps.

Given:
- Initial velocity, [tex]\(v_i = 3.15 \, \text{m/s}\)[/tex] (downward)
- Height (distance), [tex]\(h = 27.5 \, \text{m}\)[/tex]
- Acceleration due to gravity, [tex]\(a = 9.81 \, \text{m/s}^2\)[/tex] (downward)

Since the direction of the throw and gravity are both downward, we will consider downward direction as negative in our calculations.

Step-by-Step Solution:

1. Identify the kinematic equation:
The kinematic equation that relates initial velocity ([tex]\(v_i\)[/tex]), final velocity ([tex]\(v_f\)[/tex]), acceleration ([tex]\(a\)[/tex]), and distance ([tex]\(d\)[/tex]) is:
[tex]\[ v_f^2 = v_i^2 + 2ad \][/tex]

2. Substitute the known values into the equation:
[tex]\[ v_f^2 = (3.15 \, \text{m/s})^2 + 2 \cdot (9.81 \, \text{m/s}^2) \cdot (27.5 \, \text{m}) \][/tex]

3. Calculate the intermediate value [tex]\(v_f^2\)[/tex]:
[tex]\[ v_f^2 = 9.9225 + 538.55 \][/tex]
[tex]\[ v_f^2 = 548.4725 \, \text{m}^2/\text{s}^2 \][/tex]

4. Find the final velocity [tex]\(v_f\)[/tex]:
To find [tex]\(v_f\)[/tex], take the square root of [tex]\(v_f^2\)[/tex]:
[tex]\[ v_f = - \sqrt{548.4725} \][/tex]
Since the motion is downward, we take the negative value for the velocity:
[tex]\[ v_f = - 23.440829763470408 \, \text{m/s} \][/tex]

Conclusion:
The final velocity of the egg just before it hits the ground is:
[tex]\[ v_f \approx -23.44 \, \text{m/s} \][/tex]
This negative sign indicates that the velocity is indeed directed downward.