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 ball's velocity at the highest point of its trajectory is approximately 25.96 m/s.

Step-by-step solution

Find the initial velocity components in horizontal and vertical directions

We can find the initial velocity components using the given initial velocity and angle. Initial horizontal component of velocity (u_x): \(u_x = u \cos(\theta)\) Initial vertical component of velocity (u_y): \(u_y = u \sin(\theta)\) where u is the initial velocity (31.1 m/s) and θ is the angle (33.4°).

Calculate the initial horizontal component of velocity

Use the formula for the horizontal component: \(u_x = u \cos(\theta)\) \(u_x = 31.1 \times \cos(33.4^{\circ})\) \(u_x \approx 25.96 \mathrm{~m} / \mathrm{s}\) So, the initial horizontal component of the velocity is approximately 25.96 m/s.

Calculate the initial vertical component of velocity

Now, use the formula for the vertical component: \(u_y = u \sin(\theta)\) \(u_y = 31.1 \times \sin(33.4^{\circ})\) \(u_y \approx 17.02 \mathrm{~m} / \mathrm{s}\) So, the initial vertical component of the velocity is approximately 17.02 m/s.

Identify the horizontal component of the velocity at the highest point of the trajectory

As the horizontal component of the velocity remains constant throughout the entire flight, the horizontal component of the velocity at the highest point of the trajectory will remain the same, i.e., 25.96 m/s. Therefore, the horizontal component of the ball's velocity at the highest point of the trajectory is approximately \(25.96 \mathrm{~m} / \mathrm{s}\).