Problem Solving- Completing the Square. I'm not sure how to solve it!

Question

Sean and Mason run out of gas while fishing from their boat in the bay. they set off an emergency flare with an vertical velocity of 30 meters per second. The height of the flats in meters can be modeled by H(t)=-5t^2+30t, where t represents the number of seconds after launch. 

 

Sean thinks the flare shouldnreadh at least 15 meters to be seen from the shore. They want to know how long the flare will take to reach this height.

1. Write an equation

2. Simplfy function to solve it

3. Solve equation by completing the flare.

Answer 1

The flare reaches a height of 15 meters when H(t) = 15.

 

So, -5t²+30t = 15

 

      5t² - 30t + 15 = 0

 

        t² - 6t + 3 = 0

 

        t² - 6t = -3

 

        t² - 6t + 9 = -3 + 9

 

              (t-3)² = 6

 

               t-3 = ±√6

 

                  t = 3±√6 ≈ 0.55 sec or 5.45 sec  (The flare reaches a                                height of 15 m at 0.55 sec on the way up and then                              has that same height at 5.45 sec on the way down).   

 

 

Answer 2

The required height necessary to be seen, 15 meters, is stated as 5t(t) = 30t. Thus, the equation would be:

5t^2 + 30t = 15. Completing the square solution involves converting the t^2 coefficient to 1, then taking 1/2 of the t term and squaring it, then add that result to both sides of the equation, which makes the left-hand side of the equation a perfect square, then take square root of both sides of equation to get the answer, here the time in seconds.

5t^2 + 30t - 15 = 0, divide by 5 to get: t^2 + 6t -3 = 0, then t^2 + 6t = 3. Now, the t coefficient is 6, so take 1/2 of 6 to get 3, then square that to get 9. So, add 9 to both sides of equation: t^2 + 6t + 9 = 3 + 9 = 12.

Factor left side to (t + 3)(t + 3), or (t + 3)^2. So now we have:

(t + 3)^2 = 12, then t + 3 = sqrt 12 = 2 (sqrt 3). Sqrt 3 = approx. 1.732, so 2 (1.732) = 3.5 approx.

Now, t = 3.5 - 3, or 0.5 seconds. (Check: velocity of 30 meters per second for 0.5 seconds would be the 15 meters required to be seen!)