Question

A baseball is thrown with a velocity of \(31.1 \mathrm{~m} / \mathrm{s}\) at an angle of \(\theta=33.4^{\circ}\) above horizontal. What is the horizontal component of the ball's velocity at the highest point of the ball's trajectory?

Short answer

Answer: The horizontal component of the baseball's velocity at the highest point of its trajectory is approximately 26.01 m/s.

Step-by-step solution

Understand the given information

We have a baseball thrown with an initial velocity of 31.1 m/s, at an angle of 33.4 degrees above the horizontal. We need to find the horizontal component of velocity (Vx) at the highest point of the trajectory.

Calculate the horizontal component of velocity

The initial velocity can be divided into two components, horizontal (Vx) and vertical (Vy). To find the Vx, we use the cosine of the angle. Vx = V * cos(θ) where V is the initial velocity, and θ is the angle.

Plug the values and solve for Vx

Using the given initial velocity (31.1 m/s) and angle (33.4 degrees), we can now calculate Vx: Vx = 31.1 * cos(33.4°) Before solving the equation, we need to ensure that the angle is in radians. To convert degrees to radians, multiply the angle by (π/180): 33.4° * (π/180) = 0.5827 radians Now, plug in the radians into the equation: Vx = 31.1 * cos(0.5827)

Calculate the horizontal component of velocity

Now we can find the horizontal component of velocity: Vx = 31.1 * cos(0.5827) = 26.01 m/s (approx.)

Provide the answer

The horizontal component of the baseball's velocity at the highest point of its trajectory is approximately 26.01 m/s.

Key concepts

Horizontal Component of Velocity

When we talk about the horizontal component of velocity in projectile motion, we're referring to the part of an object's overall velocity that moves it along the horizontal plane. It's important to understand that this component remains constant throughout the flight of the projectile, as long as we neglect air resistance.

In our baseball example, calculating the horizontal component of velocity involves using the initial velocity and the cosine of the angle of projection. The equation is as such:
\[ V_x = V \cdot \cos(\theta) \]
where \( V_x \) is the horizontal velocity, \( V \) is the initial velocity, and \( \theta \) is the angle of projection. This is crucial to ensuring accuracy when solving for the horizontal motion of the projectile. At the peak of its flight, the only velocity the baseball retains is its horizontal component, as the vertical component momentarily becomes zero.

Trajectory of a Projectile

The trajectory of a projectile describes the path it takes through the air. This path is typically a parabolic curve when we assume uniform gravity and no air resistance. Two key components affect this trajectory: the initial velocity and the angle of projection. The horizontal component of velocity influences how far it will travel, while the vertical component affects how high it will go.

Mathematically, we depict the trajectory with the set of parametric equations for \( x(t) \) and \( y(t) \), where \( t \) is the time. These equations account for the initial velocity components in both directions and the acceleration due to gravity acting on the vertical component. In the case of our baseball, its highest point occurs when the vertical component is zero, but the horizontal motion persists, as it is unaffected by gravity.

Initial Velocity

Initial velocity, often denoted \( V \) or \( V_0 \), is the speed at which a projectile is launched or thrown. It's a vector quantity, meaning it has both magnitude and direction. The initial velocity can be resolved into two components: vertical and horizontal.

The equation for initial velocity considering the angle of projection is:
\[ V_0 = V_{x0} \cdot \cos(\theta) \]
For the baseball, the initial velocity given is \( 31.1 \text{ m/s} \), and it is launched at an angle of \( 33.4^\circ \). It's this initial speed, combined with the launch angle, that sets the stage for the entire motion of the baseball, influencing how far and how high it will go.

Angle of Projection

The angle of projection is the angle at which a projectile is launched relative to the horizontal. It is a critical factor in determining the shape and length of the projectile's trajectory. An angle of \( 45^\circ \) maximizes the range of a projectile, given a fixed initial velocity and neglecting air resistance.

In the example of the thrown baseball, the angle is \( 33.4^\circ \). The cosine of this angle is applied to calculate the horizontal component of the initial velocity, which remains constant if we neglect air resistance. On the other hand, the sine of this angle would be used if we were calculating the initial vertical component of velocity, which affects how high the baseball will rise before gravity pulls it back down.