An object is thrown upward from the top of a 10.0m high building at a speed of 10.0m/s. How fast is the object moving when it reaches the bottom of the building?

The answer is 17.2 m/s but how?

Collect your data.

you have your initial velocity which equals 10m/s
then you have your distance which is 10,
and the acceleration will be gravity which is 9.8m/s

You will then fill that information into the equation
Vf = Vi^2 + 2(a)(d)

it will look like
Vf^2 = 10^2 + 2(9.8)(10)
Vf^2 = 100 + 196

then to get rid of the exponent 2 on the Vf you square root your answer.
The square root of 296 is 17.2 m/s

Thats how it works, hope this helped.

To find the speed of the object when it reaches the bottom of the building, we can use the principles of motion and energy conservation.

First, let's break down the problem and identify the initial and final positions of the object:
- Initial position (top of the building): H_initial = 10.0 m
- Final position (bottom of the building): H_final = 0 m

We also know the initial velocity of the object, which is its speed when it is thrown upward:
- Initial velocity (upward): V_initial = 10.0 m/s

Now, let's analyze the motion of the object using the principle of energy conservation. When the object is at the top of the building, it has potential energy (PE) equal to its mass (m) multiplied by the acceleration due to gravity (g) and the height (H_initial):
PE_initial = m * g * H_initial

When the object reaches the bottom, all of its potential energy is converted into kinetic energy (KE):
KE_final = m * V_final^2 / 2

Since energy is conserved, we can equate the two expressions:
PE_initial = KE_final

Solving for V_final, we get:
V_final = sqrt(2 * g * H_initial)

Substituting the known values into the equation:
V_final = sqrt(2 * 9.8 m/s^2 * 10.0 m)
= sqrt(196 m^2/s^2)
≈ 14.0 m/s

Therefore, the object is moving at approximately 14.0 m/s when it reaches the bottom of the building.