Review Interactive LearningWare 2.2 as an aid in solving this problem. A hot air balloon is ascending straight up at a constant speed of 8.40 m/s. When the balloon is 14.0 m above the ground, a gun fires a pellet straight up from ground level with an initial speed of 27.0 m/s. Along the paths of the balloon and the pellet, there are two places where each of them has the same altitude at the same time. How far above ground level are these places

y₁ = 37.2 m,      y₂ = 22,6 m

Explanation:

For this exercise we can use the kinematic equations

For the globe, with index 1

       y₁ = y₀ + v₁ t

For the shot with index 2

       y₂ = 0 + v₂ t - ½ g t²

 At the point where the position of the two bodies meet is the same

        y₁ = y₂

        y₀ + v₁ t = v₂ t - ½ g t²

        14 + 8.40t = 27.0 t - ½ 9.8 t²

        4.9 t² - 18.6 t + 14 = 0

        t² - 3,796 t + 2,857 = 0

Let's look for time by solving the second degree equation

         t = [3,796 ±√(3,796 2 - 4 2,857)] / 2

         t = [3,796 ± 1,727] / 2

         t₁ = 2.7615 s

         t₂ = 1.03 s

Now we can calculate the distance for each time

         y₁ = v₂ t₁ - ½ g t₁²

        y₁ = 27 2.7615 - ½ 9.8 2.7615²

        y₁ = 37.2 m

        y₂ = v₂ t₂ - ½ g t₂²

        y₂ = 27 1.03 - ½ 9.8 1.03²

        y₂ = 22,612 m

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