Answer:
a) (0.0004·
- 1.55·
) m
b) 0.31 km/h·
+ 0.00008 km/h·
Step-by-step explanation:
The given parameters are;
The circular motion of the airplane = 600 km/hour
The radius of the circular path = 3 km
The time it takes the cup to reach the floor of the airplane is given as follows;
s = u·t + 1/2·g·t²
1.5 = 1/2 × 9.81 × t²
t² = 1.5/(1/2 × 9.81)
t = √(1.5/(1/2 × 9.81)) ≈ 0.553 s
The location of the plane in 0.533 s is 600 km/hour × 0.533 s ≈ 83.83 meters
The angular velocity, ω = v/r = 600 km/hour/3 km = 200°/hour
Therefore, the angle covered in 0.533 s is 0.533 s × 200°/hour ≈ 0.0296°
tan(0.0296)*3000 ≈ 1.55 m vertical
3000 - cos(0.0296)*3000 ≈ 0.0004 m horizontal
The location of the cup relative to the plane will be
0 - (-0.0004·i + 1.55·j) = 0.0004·i - 1.55·j m
b) The speed of the cup relative to the plane when it hits the floor is given by the
The speed of the plane = 600 km/hour × cos(90 - 0.0296)j + 600 km/hour × sin(90 - 0.0296)j = -0.31 km/hour·i + 599.99 km/hour·j
The speed of the cup = 600 km/hour·j
The relative speed = 600 km/hour·j - ( -0.31 km/hour·i + 599.99 km/hour·j) = 0.31 km/hour·i + 0.00008 km/hour·j
The relative speed of the cup to the plane = 0.31 km/h·i + 0.00008 km/h·j
Therefore, given that there will be no relative speed in an inertia frame, the speed of the cup relative to the plane will be higher than when dropped in an inertia frame