an aeroplane takes off at an angle of 30 degree to the horizontal. if the component of its velocity is 250 km/h, what is the actual velocity?
Yo = 250km/h = Ver. component?
Vo = 250/sin30 = 500 km/h.
To find the actual velocity of the airplane, we need to consider its velocity components along the horizontal and vertical directions.
Given:
Angle of takeoff (θ) = 30 degrees
Component of velocity (v) = 250 km/h
The horizontal component of velocity is given by: v_horizontal = v * cos(θ)
The vertical component of velocity is given by: v_vertical = v * sin(θ)
Substituting the given values, we can calculate the velocity components:
v_horizontal = 250 km/h * cos(30°)
v_horizontal = 250 km/h * (√3/2)
v_horizontal = 250 km/h * 1.732/2
v_horizontal = 433.005 km/h (approx.)
v_vertical = 250 km/h * sin(30°)
v_vertical = 250 km/h * (1/2)
v_vertical = 125 km/h
Now, to find the actual velocity, we can use the Pythagorean theorem:
velocity (v_actual) = √(v_horizontal^2 + v_vertical^2)
v_actual = √(433.005 km/h)^2 + (125 km/h)^2
v_actual = √(187744.508 km^2/h^2 + 15625 km^2/h^2)
v_actual = √(203369.508 km^2/h^2)
v_actual = 450.316 km/h (approx.)
Therefore, the actual velocity of the airplane is approximately 450.316 km/h.