Question

A baseball is launched from the bat at an angle \(\theta_{0}=30.0^{\circ}\) with respect to the positive \(x\) -axis and with an initial speed of \(40.0 \mathrm{~m} / \mathrm{s}\), and it is caught at the same height from which it was hit. Assuming ideal projectile motion (positive \(y\) -axis upward), the velocity of the ball when it is caught is a) \((20.00 \hat{x}+34.64 \hat{y}) \mathrm{m} / \mathrm{s}\) b) \((-20.00 \hat{x}+34.64 \hat{y}) \mathrm{m} / \mathrm{s}\) c) \((34.64 \hat{x}-20.00 \hat{y}) \mathrm{m} / \mathrm{s}\) d) \((34.64 \hat{x}+20.00 \hat{y}) \mathrm{m} / \mathrm{s}\).

Short answer

Answer: c) \((34.64 \hat{x}-20.00 \hat{y}) \mathrm{m} / \mathrm{s}\)

Step-by-step solution

Find the initial velocity components

We're given an initial speed of 40 m/s and an angle of \(30^\circ\) respect to the positive x-axis. To solve for the launch components, we can use simple trigonometry. The initial horizontal velocity (\(v_x\)) and vertical velocity (\(v_y\)) components can be found using the following equations: \(v_x = v_0 \cos{\theta_0}\) \(v_y = v_0 \sin{\theta_0}\) where \(v_0\) is the initial speed and \(\theta_0\) is the initial angle. \(v_x = (40 \mathrm{~m} / \mathrm{s}) \cos{30^{\circ}} = 20 \sqrt{3} \mathrm{~m} / \mathrm{s} \approx 34.64\mathrm{~m} / \mathrm{s}\) \(v_y = (40 \mathrm{~m} / \mathrm{s})\sin{30^{\circ}} = 20 \mathrm{~m} / \mathrm{s}\)

Determine the time of flight

Since it says the ball is caught at the same height as it was hit, the vertical displacement is 0. Using the vertical motion equation, we can solve for time: \(y_f = y_0 + v_{y0}t - \frac{1}{2}gt^2\) \(0 = 0 + (20\mathrm{~m} / \mathrm{s})t - \frac{1}{2}(9.8\mathrm{~m} / \mathrm{s}^2)t^2\) We can solve this equation for \(t\): \(t = \frac{2v_{y0}}{g} = \frac{2(20 \mathrm{~m} / \mathrm{s})}{9.8 \mathrm{~m} / \mathrm{s}^2} \approx 4.082\mathrm{~s}\)

Calculate the final velocity components

As the horizontal motion is not affected by gravity, the horizontal velocity remains unchanged when the ball is caught: \(v_{xf} = v_{x0} = 34.64\mathrm{~m} / \mathrm{s}\) Now let's find the vertical velocity component when the ball is caught: \(v_{yf} = v_{y0} - gt = (20 \mathrm{~m} / \mathrm{s}) - (9.8\mathrm{~m} / \mathrm{s}^2)(4.082\mathrm{~s}) \approx -20\mathrm{~m} / \mathrm{s}\)

Provide the final velocity vector

Based on the calculations, the final velocity components are 34.64 m/s in the horizontal direction and -20 m/s in the vertical direction. Combining these components to form the velocity vector: \(\vec{v_f} = (34.64 \hat{x} - 20.00 \hat{y}) \mathrm{m} / \mathrm{s}\) Thus, the correct answer is c) \((34.64 \hat{x}-20.00 \hat{y}) \mathrm{m} / \mathrm{s}\).

Key concepts

Initial Velocity Components

Understanding the initial velocity components is integral in solving problems involving projectile motion. These components are essentially the building blocks that describe how the object is moving. In physics, we often split the motion into horizontal and vertical components since these two directions are independent of each other.

To break down the initial velocity, we use trigonometry. The horizontal component ($v$_x) is found by multiplying the initial speed by the cosine of the launch angle, while the vertical component ($v$_y) is calculated by multiplying the initial speed by the sine of the launch angle. This allows us to see motion in a two-dimensional space with clear, understandable figures, simplifying the analysis of the projectile's behavior throughout its arc.

• $v$_x = v₀ × cosθ₀
• $v$_y = v₀ × sinθ₀

Time of Flight Calculation

The time of flight calculation is crucial in determining how long a projectile stays in the air. It is the duration from launch to landing when the projectile returns to its original vertical position (vertical displacement equals zero).

To find the time of flight for our baseball scenario, we used the kinematic equation which relates initial velocity, acceleration due to gravity, and displacement in the vertical direction. Since the starting and ending heights are the same and we neglect air resistance, the equation simplifies, allowing us to calculate the total time the baseball spends in the air, which is a key value for further calculations and understanding the whole picture of motion.

Vertical Displacement

Vertical displacement is the measure of the distance an object has moved in the vertical direction. For the typical projectile motion scenario where the launch and catch points are at the same elevation, the vertical displacement is zero. This simplification is helpful because it means the projectile's upward and downward movements counterbalance each other perfectly, and it links directly to the symmetry of parabolic motion. Knowing this, we were able to set the vertical displacement to zero in our kinematic equation. This principle is a cornerstone in solving a wide range of projectile problems and understanding the impact of gravity on motion.

Velocity Vector

The velocity vector combines both horizontal and vertical components of velocity, essentially telling us the speed and direction of a projectile at any given moment. At the moment the baseball in our example is caught, it has a horizontal velocity component that remains unchanged from the initial launch due to the absence of horizontal forces (assuming no air resistance). Alternatively, the vertical velocity component has changed due to acceleration under gravity.

This velocity vector is paramount in defining the final state of motion of a projectile, connecting the dots between how it started its flight and how it concludes. By keeping the horizontal constant and modifying the vertical by gravity's effect over time, we can determine the exact condition of the baseball when it lands back in a fielder's glove. Understanding this helps to map the behavior of projectiles in a clear and direct way.