A truck travels beneath an airplane that is moving 100 km/h at an angle of 49 degrees to the ground.
A) how fast must the truck travel to stay beneath the airplane? Answer in units of km/h
The answer I got was 65.6 km/h, but i just wanted to make sure someone else got the same answer.also there's a part B which i am not sure how to find...
B) What is the magnitude of the verticle component of the velocity of the plane. Answer in units of km/h
yes on a, on b, the vertical component is 100sin49deg
To solve part A, we need to break down the velocity components of both the airplane and the truck and compare them.
Let's denote the velocity of the airplane as V_airplane and the velocity of the truck as V_truck.
1. Decompose the velocity of the airplane into its horizontal (V_airplane_horizontal) and vertical (V_airplane_vertical) components:
V_airplane_horizontal = V_airplane * cos(angle)
V_airplane_vertical = V_airplane * sin(angle)
2. To stay beneath the airplane, the truck's horizontal velocity must match the horizontal velocity of the airplane, i.e.,
V_truck = V_airplane_horizontal
Given that V_airplane = 100 km/h and the angle = 49 degrees, we can calculate V_airplane_horizontal:
V_airplane_horizontal = V_airplane * cos(angle)
= 100 km/h * cos(49 degrees)
≈ 64.117 km/h
Therefore, the truck must travel at a speed of approximately 64.117 km/h to stay beneath the airplane.
Now, let's move on to part B.
To find the magnitude of the vertical component of the velocity of the plane, we use the value of V_airplane_vertical calculated earlier.
V_airplane_vertical = V_airplane * sin(angle)
= 100 km/h * sin(49 degrees)
≈ 76.286 km/h
Hence, the magnitude of the vertical component of the velocity of the plane is approximately 76.286 km/h.