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In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall with a speed of \(18 \mathrm{m} / \mathrm{s}\) at an angle of \(32^{\circ}\) above the horizontal. (a) How long does it take for the ball to reach the wall if it is \(3.8 \mathrm{m}\) away? (b) How high is the ball when it hits the wall?

Short Answer

Expert verified
(a) \(0.249\text{ s}\), (b) \(2.12\text{ m}\)

Step by step solution

01

Break Down the Initial Horizontal Motion

To find out how long it takes for the ball to reach the wall, we need to focus on the horizontal motion. The horizontal velocity \(v_x\) can be calculated using the initial speed and the angle: \(v_x = v_0 \cdot \cos(\theta)\). Plug in the given values: \(v_x = 18 \cdot \cos(32^{\circ})\). Simplifying, \(v_x \approx 15.27 \text{ m/s}\).
02

Calculate Time to Reach the Wall

The time \(t\) it takes for the ball to reach the wall can be calculated using the horizontal distance \(d = 3.8 \text{ m}\) and the horizontal velocity \(v_x\). Use the formula \(t = \frac{d}{v_x}\). Substituting the known values, \(t = \frac{3.8}{15.27}\). Solving this gives \(t \approx 0.249 \text{ s}\).
03

Break Down the Initial Vertical Motion

Now, let's find the vertical component of the velocity \(v_y\), using \(v_y = v_0 \cdot \sin(\theta)\). Substitute the values: \(v_y = 18 \cdot \sin(32^{\circ})\). Simplifying, \(v_y \approx 9.54 \text{ m/s}\).
04

Calculate the Height of the Ball at the Wall

To find the height when the ball hits the wall, use the kinematics equation for vertical motion: \( y = v_y \cdot t - \frac{1}{2} g t^2\), where \(g\) is gravity \(9.8 \text{ m/s}^2\) and initial height \(y_0 = 0\). Substitute \(v_y = 9.54\), \(t = 0.249\), and \(g = 9.8\). Solving, \( y \approx 9.54 \times 0.249 - 0.5 \times 9.8 \times (0.249)^2\) gives \( y \approx 2.12 \text{ m}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal Motion
Horizontal motion in projectile activities, like hitting a ball towards a wall, is predictable and follows a straightforward path. This motion occurs along a straight line parallel to the horizontal axis. When analyzing it, we can assume that there is no acceleration affecting the ball horizontally. To calculate the horizontal velocity, we use the formula:
  • The horizontal velocity, \(v_x\) = \(v_0 \cdot \cos(\theta)\) where \(v_0\) is the initial speed and \(\theta\) is the angle.
In our exercise, substitution of the given values yields \(v_x \approx 15.27 \text{ m/s}\). Knowing this speed is crucial because it helps us to figure out the time it takes for the object to reach a specified point. We use the formula:
  • Time, \(t = \frac{d}{v_x}\) where \(d\) is the distance to the wall.
By substituting \(d = 3.8 \text{ m}\) and \(v_x\), we find the time taken is approximately 0.249 seconds.
Vertical Motion
The vertical component of projectile motion involves movement under the influence of gravity, affecting how high or low an object travels. Unlike horizontal movement, vertical motion does involve acceleration due to gravity.The vertical velocity can be calculated as:
  • The vertical velocity, \(v_y = v_0 \cdot \sin(\theta)\) where \(v_0\) is the initial speed and \(\theta\) is the angle.
For our example, this is approximately \(9.54 \text{ m/s}\). This value is a crucial part of determining the ball's height as it moves.To find how high above the ground the object is at a particular time, the kinematics equation for vertical motion is used:
  • Height, \(y = v_y \cdot t - \frac{1}{2} g t^2\), where \(g\) is the gravitational acceleration (\(9.8 \text{ m/s}^2\)).
Solving with the time from our horizontal calculation and substituting the other known values, the height of the ball when it reaches the wall is approximately \(2.12 \text{ m}\).
Kinematics
Kinematics provides the fundamentals for understanding motion, specifically, how objects move. By studying the kinematic equations, we can predict both horizontal and vertical motions independently and calculate the exact trajectory of moving objects. Kinematic principles hinge on breaking down the initial velocities into their components:
  • Horizontal and vertical motions are treated separately.
  • The absence of horizontal acceleration simplifies calculations.
  • Vertical acceleration due to gravity affects the motion in a negatively constant manner.
These principles allow us to calculate important parameters like:
  • The time of flight using horizontal motion.
  • The height reached through vertical analysis.
Both components together provide a complete picture of the projectile's path. In scenarios like the handball exercise, understanding kinematics helps solve real-world problems like determining how fast and high an object will be at a given point in its path.

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