Question

an object is dropped from the top of a building which is 75 meters high. what's it's velocity before it reaches ground

Answer 1

Hi there! :)

`\large\boxed{38.34 m/s}`

Use the following kinematic equation to solve:

`v_(f)^(2) = v_(i)^(2) + 2ad`

Where:

vf = final velocity (?)

vi = initial velocity (0 m/s, object started at rest)

a = acceleration (gravity = 9.8 m/s)

d = distance travelled (75 m)

Plug in the given values into the equation:

`v_(f)^(2) = 0^(2) + 2(9.8)(75)`

Simplify:

`v_(f)^(2) = 2(9.8)(75)\\\\v_(f)^(2) = 1470`

Take the square root of both sides:

`v_(f) = √(1470) = 38.34 m/s`