Question

A student holds an egg outside of an open window and let's go. The window is 40 meters above the ground and the egg is falling under the acceleration of gravity, which is 9.8 m/s2. There are two equations that can be used to describe its motion over time:
How far will the egg travel after 2 seconds? Be sure to use the 5 steps to solve the problem. Show which equation you used, what each variable equals, and answer the question in a complete sentence.

Answer 1

The distance traveled by the egg after 2 seconds is 19.6 m.

The given parameters:

• Height of the window, h = 40 m

The distance traveled by the egg after 2 seconds is calculated as follows;

$h = v_0_y t + \frac{1}{2} gt^2$

where;

• $v_0_y$ is the initial vertical velocity of the egg = 0
• g is acceleration due to gravity = 9.8 m/s²
• t is the time of motion, = 2 s

$h = 0 \ + \ \frac{1}{2} \times 9.8 \times (2)^2\\\\h = 19.6 \ m$

Thus, the distance traveled by the egg after 2 seconds is 19.6 m.

Learn more about second equation of motion here: brainly.com/question/7977693