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Suppose a soccer player kicks the ball from a distance \(30 \mathrm{~m}\) toward the goal. Find the initial speed of the ball if it just passes over the goal, \(2.4 \mathrm{~m}\) above the ground, given the initial direction to be \(40^{\circ}\) above the horizontal.

Short Answer

Expert verified
The initial speed of the ball is found by determining the time it takes to reach the goal height using vertical motion and then using horizontal motion to solve for the initial speed.

Step by step solution

01

- Identify the Known Variables

We know the horizontal distance to the goal (range) is 30 m, the height of the goal is 2.4 m, and the angle of projection is 40 degrees.
02

- Break the Initial Velocity into Components

Calculate the initial velocity components. The initial speed (v) can be broken down into horizontal (vx) and vertical (vy) components using trigonometric functions. \(vx = v \cdot \cos(40^\circ)\) and \(vy = v \cdot \sin(40^\circ)\).
03

- Use Kinematic Equation for Vertical Motion

Use the kinematic equation to find the time (t) it takes for the ball to reach the height of 2.4 m: \( H = vy \cdot t - \frac{1}{2}gt^2\), where H is 2.4 m and g is 9.8 m/s^2.
04

- Solve for Time 't' in Vertical Motion

Rearrange the equation to solve for t, resulting in a quadratic equation. Substitute the known values to find the time it takes for the ball to reach 2.4 m.
05

- Use Horizontal Motion to Find Initial Speed

The horizontal range (R) is given by \( R = vx \cdot t\). Substitute for vx from step 2 and the time found in step 4 to solve for the initial speed v.

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Key Concepts

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

Initial Velocity Calculation
Understanding how to calculate initial velocity in projectile motion is crucial, as it sets the stage for the entire trajectory of the object. In the context of the soccer player's problem, where the ball needs to travel horizontally to reach a goal that is 30 meters away and also needs to pass over a 2.4-meter-tall obstacle, we need to determine the velocity with which the soccer player must kick the ball.

To start, you should note that the initial velocity is a vector, which means it has both a magnitude (the speed) and a direction. Using trigonometry, we can decompose this vector into two components: the horizontal component (vx), and the vertical component (vy). The horizontal component is responsible for the reach, while the vertical component ensures the ball will rise above the goal's height.

The equations used to calculate these components involve trigonometric functions, which link the angle of kick with the overall magnitude of the initial velocity. For a kick angle of 40 degrees, the horizontal and vertical components are found using the respective cosine and sine functions:
  • Horizontal (vx): \( vx = v \cdot \cos(40^\circ) \)
  • Vertical (vy): \( vy = v \cdot \sin(40^\circ) \)
Once both components are known, the overall magnitude of the initial velocity can be determined with the Pythagorean theorem, where \( v = \sqrt{vx^2 + vy^2} \). However, in this problem, we work backwards from the components to find the initial speed using kinematics.
Kinematic Equations
The kinematic equations are the tools that allow us to describe the motion of objects in classical physics, assuming that the acceleration is constant, as in the case of gravity in projectile motion. The vertical motion of the kicked ball can be described using the following kinematic equation:
\[ H = vy \cdot t - \frac{1}{2}gt^2 \]
where H is the height that the ball reaches (2.4 m in this example), vy is the vertical component of the initial velocity, t is the time, and g is the acceleration due to gravity (9.8 m/s²). This equation is essential because it links the time it takes for the ball to reach the height with its initial vertical speed.

Additionally, the horizontal motion can be analyzed independently as there is no acceleration in the horizontal direction if we neglect air resistance. The ball travels with a constant horizontal velocity (vx), and the range R (horizontal distance) can be given by:
\[ R = vx \cdot t \]
With the time 't' calculated from the vertical motion, one can solve for the initial speed using the equation that models the horizontal motion. By combining the motion in both directions and knowing either the horizontal range or the time of flight, we can determine the initial kinetic energy the soccer player added to the ball with his kick.
Trigonometric Functions in Physics
In physics, trigonometric functions are used to relate angles to side lengths in right triangles. They are especially relevant in projectile motion, such as calculating the initial velocity components of a soccer ball being kicked at an angle. Given an angle of projection, the trigonometric functions sine and cosine are often used to break down the initial velocity vector into its vertical and horizontal components.

The angle of projection determines how much of the initial velocity contributes to moving the soccer ball forward towards the goal, and how much lifts it off the ground to clear the goal's height. In the given problem, the initial direction is \(40^\circ\) above the horizontal:
  • For the horizontal motion component, \(vx\), we use the cosine function:

    \(vx = v \cdot \cos(\theta)\)

    Where \(\theta\) is the angle of projection, and 'v' is the magnitude of the initial velocity.
  • For the vertical motion component, \(vy\), the sine function is used:

    \(vy = v \cdot \sin(\theta)\)

Through these relationships, students can visualize the importance of angles in determining the success of kicks in soccer and other sports. Furthermore, understanding these trigonometric functions solidifies the concept of vector decomposition, which finds use in a wide array of physics problems beyond projectile motion.

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Most popular questions from this chapter

A ball is thrown horizontally from the top of a \(60.0\) -m building and lands \(100.0 \mathrm{~m}\) from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

Explain why a vector cannot have a component greater than its own magnitude.

Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither \(0^{\circ}\) nor \(90^{\circ}\) ): (a) Is the velocity ever zero? (b) When is the velocity a minimum? A maximum? (c) Can the velocity ever be the same as the initial velocity at a time other than at \(t=0 ?\) (d) Can the speed ever be the same as the initial speed at a time other than at \(t=0 ?\)

Suppose you take two steps \(\mathbf{A}\) and \(\mathbf{B}\) (that is, two nonzero displacements). Under what circumstances can you end up at your starting point? More generally, under what circumstances can two nonzero vectors add to give zero? Is the maximum distance you can end up from the starting point \(\mathbf{A}+\mathbf{B}\) the sum of the lengths of the two steps?

Derive \(R=\frac{v_{0}^{2} \sin 2 \theta_{0}}{g}\) for the range of a projectile on level ground by finding the time \(t\) at which \(y\) becomes zero and substituting this value of \(t\) into the expression for \(x-x_{0}\), noting that \(R=x-x_{0}\)

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