A soccer player kicks a ball with an initial velocity of 10m/s at an angle of 30 degrees above the horizontal. Find the; a) magnitude of the horizontal component b) Vertical component c) The magnitude of the vector.

a) V₀ₓ = 8.66 m/s

b) V₀y = 5 m/s

c) Magnitude of velocity vector = 10 m/s

Explanation:

a)

The magnitude of the horizontal component of launch velocity can easily be given by the following formula:

V₀ₓ = V₀ Cos θ

where,

V₀ₓ = horizontal component of velocity = ?

V₀ = Launch Velocity = 10 m/s

θ = Launch Angle of the ball with the horizontal = 30°

Therefore,

V₀ₓ = (10 m/s)(Cos 30°)

V₀ₓ = 8.66 m/s

b)

The magnitude of the vertical component of launch velocity can easily be given by the following formula:

V₀y = V₀ Sin θ

where,

V₀y = vertical component of velocity = ?

V₀ = Launch Velocity = 10 m/s

θ = Launch Angle of the ball with the horizontal = 30°

Therefore,

V₀y = (10 m/s)(Sin 30°)

V₀y = 5 m/s

c)

The magnitude of the velocity vector will be equal to the resultant velocity or net velocity, which is 10 m/s.

Magnitude of Velocity Vector = 10 m/s

found in the nucleus

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