SOLUTION: use the quadratic formula to determine the times when an object launched vertically from the surface of the earth at an initial speed of 40m/s reaches a height of 50 meters. Includ

The formula for the height under given conditions is

    h(t) = -5t^2 + 40t.


The coefficient at t^2 is half of the gravity acceleration with the sign "minus".

The coefficient at "t" is the given initial vertical speed of 40 m/s.


To find the time, when the object will get the height of 50 m, you should solve
this equation 

    -5t^2 + 40t = 50,

or, which is equivalent,

    t^2 - 8t + 10 = 0.


Usually, when a student obtains such a problem, it is assumed, that he (or she) just knows
the technique of solving quadratic equation using the quadratic formula

     =  =  =  = .


So, there are two solutions   =  = 1.55 seconds (rounded), which relates to ascending move,

and   =  = 6.45 seconds (rounded), which relates to the descending move.


The plot in (y,t)-coordinates is shown below.


    


        Plot  h(t) = -5t^2 + 40t (red) and  h = 50 m (green)