Question

While standing at the edge of a building's roof, Chad throws an egg upward with an initial speed of 6.00 m/s. The egg subsequently smashes on the ground, 13.0 m beneath the height from which Chad threw it. Ignore the effects of air resistance.

At what speed does the egg pass the point from which it was thrown?

Answer 1

17.0 m/s

Step-by-step explanation:

The speed of the egg as it passes the point from which it was thrown can be calculated using the equation of motion for a vertically-thrown object under constant acceleration due to gravity:

v = √(v0^2 + 2ax), where

v0 = initial speed = 6.00 m/s

a = acceleration due to gravity = 9.8 m/s^2 (downward)

x = displacement = 13.0 m (downward)

So, v = √(6.00 m/s)^2 + 2 * 9.8 m/s^2 * (-13.0 m)

= √(36 + 251.2) m^2/s^2

= √287.2 m^2/s^2

= 17.0 m/s

Therefore, the egg passes the point from which it was thrown at a speed of 17.0 m/s.

Hope this helps!