SOLUTION: 4. An airplane diving at an angle of 40.9º below the horizontal drops a mailbag from an altitude of 900 m. The bag strikes the ground 5.00 s after its release. a) What is the

Question 1177283: 4. An airplane diving at an angle of 40.9º below the horizontal drops a mailbag from an
altitude of 900 m. The bag strikes the ground 5.00 s after its release.
a) What is the speed of the plane?
b) How far does the bag travel horizontally during its fall?
c) What are the horizontal and vertical components of its velocity just before it
strikes the ground?
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
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.**

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