Problem 122 - Fundamentals of Physics

Question

You are to throw a ball with a speed of \(12.0 \mathrm{~m} / \mathrm{s}\) at a target that is height \(h=5.00 \mathrm{~m}\) above the level at which you release the ball (Fig. 4-58). You want the ball's velocity to be horizontal at the instant it reaches the target. (a) At what angle \(\theta\) above the horizontal must you throw the ball? (b) What is the horizontal distance from the release point to the target? (c) What is the speed of the ball just as it reaches the target?

Step-by-Step Solution

Answer

(a) \(\theta = 55.0^\circ\); (b) Horizontal distance = 13.9 m; (c) Speed at target = 6.88 m/s.

Step 1: Analyze the problem

To ensure the ball's velocity is horizontal when it reaches the target, it must be moving exactly horizontally at the top of its parabolic trajectory. This means the vertical component of velocity at the target must be zero.

Step 2: Determine vertical velocity component

The vertical component of the velocity when the ball is thrown is given by \( v_{0y} = v_0 \sin(\theta) \) where \( v_0 = 12.0 \, \text{m/s} \). We want \( v_{y} = 0 \) at the target. Use the kinematic equation \( v_y^2 = v_{0y}^2 - 2g(h) \) where \( g = 9.8 \, \text{m/s}^2 \) is acceleration due to gravity, to find the initial vertical velocity component.

Step 3: Solve for initial vertical velocity

Set \( v_y = 0 \) and solve for \( v_{0y} \), we have \( 0 = v_{0y}^2 - 2g(h) \). \[v_{0y} = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 5} = 9.9 \, \text{m/s}\]

Step 4: Calculate the angle \(\theta\)

Using \( v_{0y} = v_0 \sin(\theta) \), substitute \( v_{0y} = 9.9 \, \text{m/s} \) and solve for \( \theta \). \[ \sin(\theta) = \frac{v_{0y}}{v_0} = \frac{9.9}{12.0} \approx 0.825 \] \[ \theta = \arcsin(0.825) \approx 55.0^\circ \]

Step 5: Determine horizontal distance

After reaching maximum height, the ball moves horizontally at velocity \( v_{0x} = v_0 \cos(\theta) \). The time \( t \) to reach maximum height can be found using \( v_{0y} = g t \), solving for \( t \): \[ t = \frac{v_{0y}}{g} = \frac{9.9}{9.8} \approx 1.01 \, \text{s} \] The total horizontal distance \( d \) to the target is \( d = 2 v_{0x} t \).Calculate \( v_{0x} = 12.0 \cos(55.0^\circ) \approx 6.88 \, \text{m/s} \) and \[ d = 2 \cdot 6.88 \cdot 1.01 \approx 13.9 \, \text{m} \]

Step 6: Calculate the speed at the target

The speed of the ball at the target is just its horizontal speed because the vertical speed is zero. Thus, \( v = v_{0x} = 6.88 \, \text{m/s} \).

Key Concepts

Kinematic Equations

In the realm of physics, kinematic equations are the cornerstone for analyzing motion, especially when dealing with projectile motion. These equations allow us to describe the behavior of a moving object by relating its velocity, acceleration, displacement, and time. For projectile motion, the kinematic equations help us predict how an object moves under the influence of gravity when the influence of air resistance is negligible.

In this exercise, the vertical component of velocity is particularly important to ensure that when the ball reaches the target, it does so horizontally. This involves using the kinematic equation for velocity, such as:

• \( v_y^2 = v_{0y}^2 - 2g(h) \)

where \(v_y\) is the final vertical velocity, \(v_{0y}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity \(9.8 \, \text{m/s}^2\), and \(h\) is the height. Keeping \(v_y = 0\) ensures that at the peak of its trajectory, the ball has zero vertical velocity, allowing it to hit the target perfectly horizontally.

Parabolic Trajectory

A parabolic trajectory is a type of path that projectiles follow due to the influence of Earth's gravity. Initially, when an object is launched, it possesses both an upward and forward motion, resulting in a curved pathway.

For any projectile, the shape of its path can be described as a parabola. This occurs because the horizontal and vertical motions are independent of each other with gravity constantly affecting only the vertical motion. In simpler terms, while the horizontal motion is uniform (constant velocity), the vertical motion is subject to acceleration due to gravity.

Understanding the parabolic trajectory is crucial for accurately determining where the projectile will land. In this exercise, the ball thrown follows such a trajectory, rising until the vertical component of its velocity reaches zero, at which point it begins its descent, ensuring its velocity vector is horizontal as desired.

Vertical and Horizontal Components

In projectile motion, dissecting the velocity into vertical and horizontal components is essential for a clear analysis. These components allow us to predict the projectile's trajectory more accurately.

• Vertical Component (\(v_{0y}\)): This determines how high the projectile will go. It is affected by gravity, which accelerates the projectile downward, eventually bringing the vertical velocity to zero at the peak of the trajectory. The formula is: \( v_{0y} = v_0 \sin(\theta) \).
• Horizontal Component (\(v_{0x}\)): Generally unaffected by gravity, this determines how far the projectile will travel horizontally. It remains constant throughout the motion, calculated as \( v_{0x} = v_0 \cos(\theta) \).

Breaking down these components not only facilitates our calculations but also enhances understanding of how each affects the overall motion of the projectile. Together, they explain why in this exercise we can calculate the distance to the target and the time taken to reach it.

Initial Velocity

Initial velocity is the speed at which a projectile is launched. It significantly impacts the range and height of a projectile's flight. In this problem, the initial velocity \(v_0\) is given as \(12.0 \, \text{m/s}\).

The importance of initial velocity lies in its division into vertical and horizontal components, which determine the projectile's path. A higher initial velocity increases the projectile's reach both vertically and horizontally, depending on the angle of projection.

The initial angle \(\theta\), obtained through the sine function, dictates the proportion of the velocity allocated between these two components. The solution to the exercise shows that throwing the ball at an angle of approximately \(55.0^\circ\) above the horizontal allows for optimal distribution to meet the conditions of the exercise, with the projectile traveling the desired distance and ensuring its velocity is horizontal at the target.