A ball is kicked with an initial velocity of 20.0m/s at an angle of 30.0 degrees above the ground. What are the horizontal and the vertical components of its velocity after 1 second?

Answers

A ball

To find the horizontal and vertical components of velocity after 1 second, we need to consider the equations of motion and break down the initial velocity into its components.

The initial velocity (v₀) has two components: one in the horizontal direction (v₀x) and one in the vertical direction (v₀y).

The horizontal component (v₀x) remains constant since there is no acceleration acting in the horizontal direction. Therefore, v₀x = v₀ * cos(θ), where θ is the angle above the ground.

Substituting the values, v₀ = 20.0 m/s and θ = 30.0 degrees, we can solve for v₀x:

v₀x = 20.0 m/s * cos(30.0 degrees)
= 20.0 m/s * 0.866
≈ 17.32 m/s

So, the horizontal component of the velocity is approximately 17.32 m/s after 1 second.

On the other hand, the vertical component (v₀y) will be affected by the gravitational acceleration (-9.8 m/s²). Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can find v₀y after 1 second.

v₀y = u + at
v₀y = 0 m/s + (-9.8 m/s²) * 1 s
v₀y = -9.8 m/s

The negative sign indicates that the ball is moving downward.

Therefore, after 1 second, the horizontal component of the velocity remains approximately 17.32 m/s, while the vertical component of the velocity changes to -9.8 m/s.