Question 1177283
Let's solve this problem step-by-step.

**Given Information:**

* Angle (θ): 40.9° below the horizontal
* Altitude (y): 900 m
* Time (t): 5.00 s

**a) What is the speed of the plane?**

1.  **Vertical Motion:**
    * We can use the kinematic equation: y = v₀y * t + (1/2) * g * t²
    * Where:
        * y = vertical displacement (900 m)
        * v₀y = initial vertical velocity
        * g = acceleration due to gravity (9.8 m/s²)
        * t = time (5.00 s)

2.  **Calculate Initial Vertical Velocity (v₀y):**
    * 900 = v₀y * 5 + (1/2) * 9.8 * 5²
    * 900 = 5v₀y + 4.9 * 25
    * 900 = 5v₀y + 122.5
    * 777.5 = 5v₀y
    * v₀y = 155.5 m/s

3.  **Calculate Speed of the Plane (v₀):**
    * v₀y = v₀ * sin(θ)
    * v₀ = v₀y / sin(θ)
    * v₀ = 155.5 / sin(40.9°)
    * v₀ = 155.5 / 0.6549
    * v₀ ≈ 237.44 m/s

**b) How far does the bag travel horizontally during its fall?**

1.  **Calculate Initial Horizontal Velocity (v₀x):**
    * v₀x = v₀ * cos(θ)
    * v₀x = 237.44 * cos(40.9°)
    * v₀x = 237.44 * 0.7558
    * v₀x ≈ 179.46 m/s

2.  **Calculate Horizontal Distance (x):**
    * x = v₀x * t
    * x = 179.46 * 5.00
    * x ≈ 897.3 m

**c) What are the horizontal and vertical components of its velocity just before it strikes the ground?**

1.  **Horizontal Velocity (vx):**
    * vx = v₀x (horizontal velocity remains constant)
    * vx ≈ 179.46 m/s

2.  **Vertical Velocity (vy):**
    * vy = v₀y + g * t
    * vy = 155.5 + 9.8 * 5
    * vy = 155.5 + 49
    * vy = 204.5 m/s

**Answers:**

* **a) The speed of the plane is approximately 237.44 m/s.**
* **b) The bag travels approximately 897.3 m horizontally.**
* **c) The horizontal component of the velocity is approximately 179.46 m/s, and the vertical component is 204.5 m/s.**