Question

A student falls off a cliff into the lake 54.0 m below. What is the final velocity of the student?

Answer 1

`v_(y)` = -32.53 m / s

this velocity is directed downwards

Step-by-step explanation:

This is a free fall exercise, let's use the expression

`v_(y)^(2)` = v_{oy}^{2} + 2 g (y -yo)

where we are assuming that there is friction with the air, as the body falls its initial velocity is zero

v_{oy} = √ 2g (y - y₀)

let's calculate

v_{y} = √ (2 9.8 (0-54.0))

`v_(y)` = -32.53 m / s

this velocity is directed downwards