Problem 46 - Fundamentals of Physics

Question

In basketball, hang is an illusion in which a player seems to weaken the gravitational acceleration while in midair. The illusion depends much on a skilled player's ability to rapidly shift the ball between hands during the flight, but it might also be supported by the longer horizontal distance the player travels in the upper part of the jump than in the lower part. If a player jumps with an initial speed of \(v_{0}=7.00 \mathrm{~m} / \mathrm{s}\) at an angle of \(\theta_{0}=35.0^{\circ}\), what percent of the jump's range does the player spend in the upper half of the jump (between maximum height and half maximum height \() ?\)

Step-by-Step Solution

Answer

The player spends approximately 51.2% of the jump's range in the upper half.

Step 1: Calculate the Time of Flight

To find the total time the player spends in the air, we use the formula for vertical motion: \[t_f = \frac{2v_0 \sin \theta_0}{g}\]where \(g = 9.81 \, \text{m/s}^2\) is the acceleration due to gravity. Substituting the given values, \(v_0 = 7.00 \, \text{m/s}\) and \(\theta_0 = 35.0^\circ\), we calculate:\[t_f = \frac{2 \times 7.00 \times \sin(35.0^\circ)}{9.81} \approx 0.821 \, \text{s} \]

Step 2: Determine Maximum Height

The maximum height \( H \) can be calculated by the formula:\[ H = \frac{(v_0 \sin \theta_0)^2}{2g} \]Substituting the values, we get:\[ H = \frac{(7.00 \sin(35.0^\circ))^2}{2 \times 9.81} \approx 1.208 \, \text{m} \]

Step 3: Find Half Maximum Height Time

The vertical position when at half maximum height \(H/2\) can be determined using:\[ \frac{v_0 \sin(\theta_0)}{g}t - \frac{1}{2}gt^2 = \frac{1.208}{2} \]This equation is quadratic in form \(at^2 + bt + c = 0\) where \( a = -\frac{g}{2} \, b = v_0 \sin(\theta_0), \text{ and } c = -\frac{1.208}{2} \). Solving gives two values, where the second represents the ascent to half height.

Step 4: Solve the Quadratic Equation for Half Height Time

Let's compute the time using the quadratic equation:\[ a = -4.905 \,, \; b = 4.01 \,, \; c = -0.604 \\implies t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Calculate the roots and find the time for when the player reaches \( H/2 \) during ascent:\[ t_{H/2} = \frac{-4.01 \pm \sqrt{(4.01)^2 - 4 \times (-4.905) \times (-0.604)}}{2 \times (-4.905)} \\approx 0.42 \, \text{s} \]

Step 5: Calculate Time Above Half Maximum Height

The time spent above half the maximum height \( H/2 \) is what remains from half height to the peak and then symmetrically back above \( H/2 \) to descent, which takes twice the time from \( H/2 \) to \( H \):\[ t_{\text{top half}} = 0.821 - 2(0.42) \approx 0.42 \, \text{s} \]

Step 6: Calculate Percentage of Time in Upper Half

Calculate the percentage of time spent in the upper part of the jump by comparing it to the total time:\[ \text{Percentage} = \left(\frac{0.42}{0.821}\right) \times 100 \approx 51.2\% \]

Key Concepts

Gravitational Acceleration

Gravity is the force that pulls objects toward the center of the Earth. This force is responsible for the acceleration experienced by objects in free fall or when they are thrown into the air. In physics, we denote gravitational acceleration by the symbol \( g \), and on Earth, its value is approximately \( 9.81 \, \text{m/s}^2 \).
Understanding gravitational acceleration is crucial because it influences every aspect of projectile motion. When a player jumps, gravity is the only force acting against their upward motion, causing them to eventually slow down, stop, and reverse direction back to the ground.

• Gravitational acceleration remains constant near the Earth's surface and affects the time, height, and distance of any projectile.
• During a jump, gravity causes the player to experience a deceleration upward and acceleration downward with the same magnitude.

Time of Flight

The time of flight in projectile motion is the total time an object spends in the air from when it is launched until it returns to the same vertical level. Calculating this time is essential for understanding the entire duration of the jump.
To calculate the time of flight, you can use the formula: \[ t_f = \frac{2v_0 \sin \theta_0}{g}\]Where:

• \( v_0 \) is the initial velocity.
• \( \theta_0 \) is the launch angle.
• \( g \) is the gravitational acceleration.

In our exercise, substituting the given values led to a time of flight of approximately 0.821 seconds.
This calculated time reflects how long the entire jump takes and helps to further calculate specific parts of the motion, like time reaching certain heights.

Maximum Height

When an object is projected upward, the maximum height is the highest point it reaches before starting to descend. This is a significant aspect of projectile motion, as it gives insights into the jump's peak performance.
The maximum height can be calculated using the formula:\[H = \frac{(v_0 \sin \theta_0)^2}{2g}\]Where the terms denote the same values as mentioned before.

• It provides a measurement of how high the player can jump.
• The maximum height in our exercise turned out to be about 1.208 meters.

Using this height, other calculations can be analyzed, like time spent above certain positions. It highlights the symmetry of projectile motion: time to ascend equals time to descend.

Quadratic Equation in Physics

Quadratic equations frequently appear in physics problems involving motion, especially when solving for time or distance in projectile scenarios. In this exercise, a quadratic equation is used to find the time at which the player reaches half of the maximum height.
The quadratic equation formula:\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Allows us to solve for time, where \(a, b,\) and \(c\) are constants derived from rearranging kinematic equations.
For our calculation:

• \(a = -\frac{g}{2} \)
• \(b = v_0 \sin(\theta_0)\)
• \(c = -\frac{H}{2}\)

Solving this quadratic equation gives the critical times when the height equals half of the jump. The process involves either deriving the physical characteristics or directly using observed data to form standard equations.