A bomb is dropped out of a plane 100 m above the ground, how fast will it be moving when it hits the ground?

Vf^2=vi2+2ad

Vf^2=2(-9.8)(100)
Vf^2=1960
Vf=44.3m/s down

To calculate the speed at which the bomb will hit the ground, we can use the laws of motion. The bomb is subject to both gravitational acceleration and air resistance, so we'll assume that air resistance is negligible for simplicity.

First, let's assume the bomb is not initially moving when dropped from the plane. The initial velocity, u, would be zero. We can use the following equation to determine the final velocity, v, when the bomb hits the ground:

v^2 = u^2 + 2as

Where:
v = final velocity
u = initial velocity
a = acceleration
s = distance

In this case, the initial velocity is 0, and the bomb is subject to the acceleration due to gravity, which is approximately 9.81 m/s^2 downward. The distance fallen, s, is 100 m. Plugging these values into the equation, we get:

v^2 = 0 + 2(-9.81)(100)

Simplifying:

v^2 = -1962

Now, we take the square root of both sides to find the final velocity:

v = √(-1962)

Since we are dealing with physical quantities, a negative square root doesn't make sense in this context. Therefore, the bomb will hit the ground at a speed of approximately 44.26 m/s.

Please note that this calculation assumes no air resistance and a constant acceleration due to gravity. In reality, air resistance and other factors could affect the speed of the falling bomb.