Question 1127955
{{{s=4.9t^2+v[0]t}}}

a) A bolt falls off an airplane at an altitude of {{{500m}}}. Approximately how long does it take the bolt to reach the​ ground? 

{{{s=4.9t^2 + v[0]t}}} where 
{{{s}}} = the distance above ground (the height or altitude), 
{{{t }}}= time in seconds, and 
{{{v[0]}}} is the initial velocity (zero in this case).

{{{4.9t^2 = 500}}} Divide both sides by {{{4.9}}}
{{{t^2 = 500/4.9}}} 
{{{t^2 = 102.04082}}} 
{{{t = sqrt(102.04082)}}}
{{{t = 10.10s}}}


b) A ball is thrown downward at a speed of {{{40(m/s)}}} from an altitude of {{{500m}}}. Approximately how long does it take the ball to reach the ground? 

Always assume positive = up and negative = down.
So the acceleration of gravity is *always* negative:

{{{a=-9.81(m/s^2)}}} (acceleration due to gravity)
In your problem, the object lands {{{500m}}} below the starting point, so 

{{{s=-500}}}
{{{v[i])=40(m/s)}}}

Use this equation:

{{{s=v[i])t+(1/2)a*t^2}}}
{{{-500=40(m/s)t+(1/2)(-9.81(m/s^2))*t^2}}}....we have to find the time {{{t}}}
{{{-500=40(m/s)t-4.905(m/s^2)*t^2}}}
{{{4.905(m/s^2)*t^2-40 (m/s)t-500=0}}}
{{{t}}} ≈ {{{14.9661s}}}


c) Approximately how far will an object fall in {{{6}}}sec, when thrown downward at an initial velocity of {{{40(m/s)}}} from a plane? 

{{{v[i])=40 ​(m/sec)}}}
{{{t=6s}}}
{{{s= 40*6+(1/2)9.81*6^2}}}..............{{{a=g=9.81(m/s^2)}}}
{{{s=240 +176.58}}}
{{{s=416.58m}}}