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?

Question

The answer is 17.2 m/s but how?

Answer 1

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.