Question

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?

Answer 1

Answer:

1.42 s

Explanation:

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

$s=\frac{1}{2}at^2$

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

$s=10.04 t^2$

this means that

$\frac{1}{2}g = 10.04$

so the acceleration of gravity on the body is

$g=2\cdot 10.04 = 20.08 m/s^2$

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

$v=gt$

In this case,

g = 20.08 m/s^2

So the time taken to reach a velocity of

v = 28.6 m/s

is

$t=\frac{v}{g}=\frac{28.6 m/s}{20.08 m/s^2}=1.42 s$