A hot air balloon is ascending straight up at a constant speed of 5.10 m/s. When the balloon is 11.0 m above the ground, a gun fires a pellet straight up from ground level with an initial speed of 39.0 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

.y₂ = 0.5704 m ,    y2= 44.47 m,

Explanation:,

For this exercise we will use the kinematics relations

Ball

        y₁ = y₀ + v₀₁ t

        y₀ = 11.0 m

        v₀₁ = 5.10 m / s

Pellet

        y₂ = 0 + v₀₂ t - ½ g t²

        V₀₂ = 39.0 m / s

At the meeting point the two bodies have the same height

            y₁ = y₂

           y₀ + v₀₁ t = v₀₂ t -1/2 g t²

            11 + 5.1 t = 39 t - ½ 9.8 t²

           4.9 t² - 33.9 t +11 = 0

           .t2 - 6,918 + 2,245 = 0

             t = [6,918 ±√ 6,918 2 - 4 2,245)] / 2

             t = [6,918 ± 6.2356.] / 2

             t₁ = 6.58 s

             t₂ = 0.3412 s

Let's calculate the positions for each time

        t₂ = 0.3412 s

              y₂ = 39 t - ½ 9.8 t2

              y₂ = 39 0.3112 - ½ 9.8 0.3412²

             y₂ = 0.5704 m

    .t1 = 6.58 s

             .₂ = 39 6.58 - ½ 9.8 6.58 ^ 2

              y2= 44.47 m

we need to know the time interval during which the force acts. impulse of force = momentum change

it's a b and