An airplane takes off a runway at a constant speed of 49m/s at constant angle 30 to the horizontal
1 answer:
Complete Question
An airplane takes off a runway at a constant speed of 49 m/s at constant angle 30 to the horizontal.How high (in meters ) is the airplane above the ground 13 seconds after takeoff?
Answer:
The height is
Explanation:
From the question we are told that
The speed at which the plane takes off is
The angle at which it takes off is
The time taken is
The vertical distance traveled is mathematically represented as
Substituting values
You might be interested in
Answer:
a) - 72.5°c
b) pressure = 3625.13 Pa
c) density = 0.063 kg/m^3
d) it is a subsonic aircraft
Explanation:
a) Determine Temperature
Temperature at 19.5 km ( 19500 m )
T = -131 + ( 0.003 * altitude in meters )
= -131 + ( 0.003 * 19500 ) = - 72.5°c
b) Determine pressure and density at 19.5 km altitude
Given :
Po (atmospheric pressure at sea level ) = 101kpa
R ( gas constant of air ) = 0.287 KJ/Kgk
T = -72.5°c ≈ 200.5 k
pressure = 3625.13 Pa
hence density = 0.063 kg/m^3
attached below is the remaining part of the solution
C) determine if the aircraft is subsonic or super sonic
Velocity ( v ) = =
= 283.8 m/s
hence it is a subsonic aircraft
Answer:
a) 17.49 seconds
b) 13.12 seconds
c) 2.99 m/s²
Explanation:
a) Acceleration = a = 1.35 m/s²
Final velocity = v = 85 km/h =
Initial velocity = u = 0
Equation of motion
Time taken to accelerate to top speed is 17.49 seconds.
b) Acceleration = a = -1.8 m/s²
Initial velocity = u = 23.61\ m/s
Final velocity = v = 0
Time taken to stop the train from top speed is 13.12 seconds
c) Initial velocity = u = 23.61 m/s
Time taken = t = 7.9 s
Final velocity = v = 0
Emergency acceleration is 2.99 m/s² (magnitude)