Answer:
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
