Question

A baseball is thrown at a 2.25-degree angle with an initial velocity of 65 meters per second. Assuming no air resistance, what is the horizontal component of the ball's velocity?

Answer 1

To find the horizontal component of a baseball's velocity, thrown at a 2.25-degree angle with an initial velocity of 65 meters per second, use the cosine function: Vx = V * cos(theta). The calculation gives an approximate value of 64.96 m/s for the horizontal velocity component.

Step-by-step explanation:

The horizontal component of the ball's velocity can be found using the cosine function of the angle, since you are looking for the adjacent side of the right-angle triangle formed by the velocity vector. Given the angle θ=2.25° and the initial velocity v=65 m/s, you use the equation:

vx = v × cos(θ)

Plugging in the values, you get:

vx = 65 m/s × cos(2.25°)

The horizontal component (vx) is approximately:

vx ≈ 64.96 m/s

This value is very close to the original velocity since the angle is very small, which means most of the ball's velocity is in the horizontal direction.