An aeroplane takes off from runway making an angle of 30 horizontally at a speed 500ms find the height attained by the aeroplane after 15s?
Vertical velocity (component) of the plane= 500sin30° m/s= 250 m/s.
H= ut (REMEMBER! IT IS NOT A PROJECTILE! IT IS A PLANE AND IT WILL MOST PROBABLY RISE WITH THE SAME SPEED IN THIS SCENARIO!).
H= ut
H= 250*15
H= 3750m. (Height attained by it after 15s).
h/(500*15) = 1/2
Review the properties of a 30-60-90 right triangle.
To find the height attained by the airplane after 15 seconds, we can use trigonometry and basic kinematic equations.
Given:
Angle of elevation (θ) = 30 degrees
Initial speed (V₀) = 500 m/s
Time (t) = 15 seconds
Step 1: Resolve the Initial Speed
Since the initial speed is given at an angle to the horizontal, we need to resolve it into its horizontal and vertical components.
Horizontal component (Vx) = V₀ * cos(θ)
Vertical component (Vy) = V₀ * sin(θ)
Using the given values:
Vx = 500 * cos(30)
Vy = 500 * sin(30)
Step 2: Calculate the Vertical Displacement
The vertical displacement represents the change in height attained during the given time.
Vertical displacement (Δy) = Vy * t + (0.5) * g * t^2
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Using the given values:
Δy = (500 * sin(30)) * 15 + (0.5) * 9.8 * (15)^2
= 7500 + 1102.5
= 8602.5 meters
Therefore, the height attained by the airplane after 15 seconds is approximately 8602.5 meters.