The equation for free fall at the surface of a celestial body in outer space (s in meters, t in seconds) is sequals10.04tsquared. how long does it take a rock falling from rest to reach a velocity of 28.6 startfraction m over sec endfraction on this celestial body in outer space?

1.42 s

Explanation:

The equation for free fall of an object starting from rest is generally written as

where

s is the vertical distance covered

a is the acceleration due to gravity

t is the time

On this celestial body, the equation is

this means that

so the acceleration of gravity on the body is

The velocity of an object in free fall starting from rest is given by

In this case,

g = 20.08 m/s^2

So the time taken to reach a velocity of

v = 28.6 m/s

is

answer: a. same unit

explanation: