Question

Scenario

Answer 1

Answer:

 t = 23.255 s,   x = 2298.98 m,    v_y = - 227.90 m / s

Explanation:

After reading your extensive writing, we are going to solve the approach.

The initial speed of the plane is 250 miles / h and it is at an altitude of 2650 m; In general, planes fly horizontally for launch, therefore this is the initial horizontal speed.

As there is a mixture of units in different systems we are going to reduce everything to the SI system.

         v₀ₓ = 250 miles h (1609.34 m / 1 mile) (1 h / 3600 s) = 111.76 m / s

         y₀ = 2650 m

Let's set a reference system with the x-axis parallel to the ground, the y-axis is vertical. As time is a scalar it is the same for vertical and horizontal movement

Y axis  

       y = y₀ + v₀ t - ½ g t²

the initial vertical velocity when the cargo is dropped is zero and when it reaches the floor the height is zero

       0 = y₀ + 0 - ½ g t²

       t = $\sqrt{ \frac{2 y_o}{g} }$

       t = √(2 2650/ 9.8)

       t = 23.255 s

Therefore, for the cargo to reach the desired point, it must be launched from a distance of

       x = v₀ₓ t

       x = 111.76 23.255

       x = 2298.98 m

at the point and arrival the speed is

        vₓ = v₀ₓ = 111.76

     

vertical speed is

         v_y = v_{oy} - gt

          v_y = 0 - gt

          v_y = - 9.8 23.25 555

         v_y = - 227.90 m / s

the negative sign indicates that the speed is down

in the attachment we have a diagram of the movement