Given a helicopter flying up and sideways at a velocity of 95 m/s at an angle of 35º to the ground, find the horizontal component of the velocity.

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;

Where;

vₓ = v × cos(θ) = The horizontal component

= 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.

there are no contact forces mentioned on the list of choices that you've provided.

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