Respuesta :
Answer:
The horizontal component of the velocity of the helicopter is approximately 77.82 m/s
Explanation:
The given parameters are;
The speed of the helicopter, v = 95 m/s
The angle in which the helicopter is flying = 35° to the ground
The horizontal component of the velocity is given by the component of the magnitude, R, of the velocity of the helicopter in the x-direction as follows;
[tex]v = \sqrt{v_x^2 + v_y^2}[/tex]
Where;
vₓ = v × cos(θ) = The horizontal component
[tex]v_y = v \times sin(\theta)[/tex] = The vertical component
∴ The horizontal component of the velocity, vₓ, of the helicopter with 95 m/s velocity moving at an angle of 35° to the horizontal is given as follos
vₓ = 95 × cos(35°) ≈ 77.82 m/s
The horizontal component of the velocity of the helicopter ≈ 77.82 m/s.
