Question

Neglect air resistance for the following. A soccer ball is kicked from the ground into the air. When the ball is at a height of \(12.5 \mathrm{~m}\), its velocity is \((5.60 \hat{x}+4.10 \hat{y}) \mathrm{m} / \mathrm{s}\) a) To what maximum height will the ball rise? b) What horizontal distance will be traveled by the ball? c) With what velocity (magnitude and direction) will it hit the ground?

Short answer

Question: Find the maximum height (a), the horizontal distance (b), and the velocity (magnitude and direction) the ball hits the ground with (c) when a soccer ball is kicked into the air with an initial position of (0, 12.5) m and an initial velocity of (5.60, 4.10) m/s. Answer (a): The maximum height of the ball is approximately [calculated value] m. Answer (b): The horizontal distance traveled by the ball is approximately [calculated value] m. Answer (c): The velocity vector when the ball hits the ground has a magnitude of approximately [calculated value] m/s and a direction of approximately [calculated value] degrees from the horizontal axis.

Step-by-step solution

Part (a) - Maximum Height

First, let's find the initial velocity in the \(y\) (vertical) direction. We're given the velocity when the ball is at a height of \(12.5 \mathrm{m}\), which is \((5.60 \hat{x}+4.10 \hat{y}) \mathrm{m} / \mathrm{s}\). The \(y\) component of this velocity vector is \(4.10 \mathrm{m/s}\). Now, the ball reaches its maximum height when \(v_y = 0\). Using equation 3, we can find the time \(t\) when the ball reaches its maximum height: \(0 = v_{0y} - gt\) \(t = v_{0y}/g\) Now we can use equation 4 to find the height \(y_{\text{max}}\) at that time: \(y_{\text{max}} = y_0 + v_{0y}t - \frac{1}{2}gt^2 = y_0 + v_{0y}\frac{v_{0y}}{g} - \frac{1}{2}g\left(\frac{v_{0y}}{g}\right)^2\) By substituting the given value for the height \((12.5 \mathrm{m})\) and velocity (\(4.10 \mathrm{m/s}\)) at that height, we can calculate the maximum height \(y_{\text{max}}\).

Part (b) - Horizontal Distance

To find the horizontal distance traveled by the ball, we need to find the time \(t\) the ball spends in the air before returning to the ground. First, we can find the initial velocity in the \(x\) (horizontal) direction. We know the velocity when the ball is \(12.5 \mathrm{m}\) above the ground is \((5.60 \hat{x}+4.10 \hat{y}) \mathrm{m/s}\). The \(x\) component of this velocity is \(5.60 \mathrm{m/s}\), which is the initial horizontal velocity, \(v_{0x}\), since there is no acceleration in the \(x\) direction. Next, we can find the time \(t\) when the ball returns to the ground. The ball returns to the ground when \(y = 0\). Using equation 4, we can find the time \(t\) when the height is \(0\): \(0 = y_0 + v_{0y}t - \frac{1}{2}gt^2\) Now that we have the time \(t\), we can calculate the horizontal distance \(x\) using equation 2: \(x = x_0 + v_{0x}t\) By substituting the initial horizontal velocity (\(5.60 \mathrm{m/s}\)) and the time \(t\), we can calculate the horizontal distance \(x\).

Part (c) - Velocity When Hitting the Ground

We found the time \(t\) when the ball returns to the ground in part (b). We can now determine the vertical velocity component \(v_y\) when the ball hits the ground using equation 3: \(v_y = v_{0y} - gt\) We already know that the horizontal component of the velocity \(v_x\) is \(5.60 \mathrm{m/s}\), as it remains constant during the motion. Finally, we can find the magnitude and direction of the velocity vector when the ball hits the ground. To find the magnitude of the velocity, we can use the Pythagorean theorem: \(v_{\text{ground}} = \sqrt{v_x^2 + v_y^2}\) To find the direction of the velocity vector (angle \(\theta\) with respect to the \(x\)-axis), we can use the inverse tangent function: \(\theta = \arctan\left(\frac{v_y}{v_x}\right)\) After finding the magnitude and direction of the velocity, we can express the final answer as a vector in terms of its magnitude and direction.

Key concepts

Maximum Height of a Projectile

Understanding the peak altitude reached by a projectile involves analyzing its vertical motion. When a soccer ball is kicked into the air, it ascends until gravity reduces its upward speed to zero. At this instant, known as the climax of its trajectory, the ball attains its maximum height.

To calculate this height for the soccer ball mentioned in the exercise, we use the initial vertical speed, which is the y-component of the velocity vector (4.10 m/s), and the acceleration due to gravity (approximately 9.81 m/s² downwards). The equation \(y_{\text{max}} = y_0 + v_{0y}\frac{v_{0y}}{g} - \frac{1}{2}g\left(\frac{v_{0y}}{g}\right)^2\) combines these values to yield the maximum height above the starting point. For effective learning, visualizing this peak as the apex of a parabola can be helpful.

Horizontal Distance Calculation

The distance a projectile travels horizontally, known as its range, is determined by its initial horizontal velocity and the time it remains airborne. There is no horizontal acceleration because air resistance is ignored, so the horizontal velocity stays constant.

For the kicked soccer ball, the horizontal component (5.60 m/s) indicates the constant speed at which it covers ground. After determining the time the ball is in flight, you can apply the equation \(x = x_0 + v_{0x}t\) to find the overall horizontal distance. Remember to enhance comprehensibility by relating these calculations to everyday experiences like kicking a ball or throwing an object.

Velocity Vector Components

Any velocity in projectile motion can be dissected into x (horizontal) and y (vertical) components. These components allow the description of the motion in two dimensions independently. For instance, the soccer ball's velocity when reaching 12.5 m is specified as \(5.60 \hat{x}+4.10 \hat{y}\) m/s.

Through illustrations or interactive tools, visualize the vector as an arrow where horizontal and vertical components represent its width and height, showing that the actual path of the projectile is a combination of these two motions. The inclusion of these visual aids will augment understanding significantly.

Kinematic Equations

Kinematic equations mathematically describe the motion of objects. In projectile motion, these equations help calculate time, velocity, and displacement in terms of the acceleration due to gravity.

Adopting equations like \(v_y = v_{0y} - gt\) and \(y = y_0 + v_{0y}t - \frac{1}{2}gt^2\), students can solve for various unknowns. For the soccer ball in our example, these equations are pivotal for determining its maximum height, the time to fall back to the ground, and its impact velocity. When teaching, emphasize the importance of understanding these formulas rather than memorizing them, encouraging a methodical approach to problem-solving.