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

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

15.3 cos 65 degrees trig