Problem 86 - Physics for Scientists and Engineers with Modern Physics

Question

A basketball leaves a player's hands at a height of \(2.10 \mathrm{~m}\) above the floor. The basket is \(3.05 \mathrm{~m}\) above the floor. The player likes to shoot the ball at a \(38.0^{\circ}\) angle. If the shot is made from a horizontal distance of \(11.00 \mathrm{~m}\) and must be accurate to \(\pm 0.22 \mathrm{~m}\) (horizontally), what is the range of initial speeds allowed to make the basket?

Step-by-Step Solution

Answer

The range of initial speeds allowed is approximately 9.96 m/s to 10.22 m/s.

Step 1: Determine Projectile Motion Equations

We have a projectile problem where the basketball is a projectile with initial speed \( v_0 \) and angle \( \theta = 38.0^\circ \). We need the equations for horizontal motion \( x = v_0 \cos \theta \cdot t \) and vertical motion \( y = v_0 \sin \theta \cdot t - \frac{1}{2} g t^2 \), where \( g = 9.81 \text{ m/s}^2 \). The initial vertical position is \( y_0 = 2.10 \text{ m} \). For the basket, \( y = 3.05 \text{ m} \).

Step 2: Substitute into Vertical Equation

Substitute the known quantities into the vertical motion equation to find \( t \):\[ 3.05 = 2.10 + (v_0 \sin 38.0^\circ) \cdot t - \frac{1}{2} \cdot 9.81 \cdot t^2 \].Simplify this to find a quadratic in \( t \) when \( v_0 \) is determined.

Step 3: Solve for Time Intervals

The quadratic equation from Step 2 needs to be solved for \( t \), given the vertex is the basket height difference to help identify ground time of the basket shot: \( 0.95 = (v_0 \sin 38.0^\circ) \cdot t - \frac{1}{2} \cdot 9.81 \cdot t^2 \). Use the quadratic formula to solve: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Step 4: Incorporate Horizontal Range Limitations

For horizontal motion, the ball's range is \( x = 11.00 \pm 0.22 \text{ m} \). Use \( x = v_0 \cos 38.0^\circ \cdot t \) for the distance. Set the two equations \( x_{min} = v_0 \cos 38.0^\circ \cdot t \) and \( x_{max} = v_0 \cos 38.0^\circ \cdot t \) where \( x_{min} = 10.78 \text{ m} \) and \( x_{max} = 11.22 \text{ m} \).

Step 5: Solve for Initial Speed

Solve the inequalities for allowed \( v_0 \) using: \[ v_0 = \frac{x}{\cos 38.0^\circ \cdot t} \] with \( x_{min} \) and \( x_{max} \). Evaluate to find values satisfying both vertical and horizontal motion constraints.

Key Concepts

Initial Speed Determination

Understanding how to determine the initial speed in projectile motion is crucial for solving various motion problems like shooting a basketball. The initial speed, often denoted as \(v_0\), is the speed at which a projectile is launched. It's essential because it affects both the horizontal and vertical components of the motion.
To find \(v_0\), you must consider both the projectile's angle of launch and the subsequent motion equations. Knowing that the basketball is shot at a \(38.0^{\circ}\) angle, we use trigonometric relationships:

• The horizontal component: \(v_0 \cos \theta\).
• The vertical component: \(v_0 \sin \theta\).

These components help break down the motion into manageable parts and allow us to calculate the speed needed to achieve specific objectives, such as reaching the target basket height and distance.

Horizontal Motion

Horizontal motion in projectile motion problems, like shooting a basketball, is simplified because there is no acceleration in the horizontal direction (assuming air resistance is negligible).
The relationship is expressed as \(x = v_0 \cos \theta \cdot t\). Here, \(x\) is the horizontal distance the ball travels, \(v_0 \cos \theta\) is the horizontal component of the initial speed, and \(t\) is the time in the air.
For this basketball problem, the goal is for the ball to travel between \(10.78 \text{ m}\) and \(11.22 \text{ m}\). To find the initial speed compatible with these distances, we substitute these values into the horizontal motion equation and solve for \(v_0\). This gives us the range of initial speeds allowed to ensure the ball lands accurately at the desired horizontal position.

Vertical Motion

Vertical motion involves the effects of gravity and the initial vertical speed when dealing with projectile motion.
The vertical motion of the basketball can be described by the equation: \(y = v_0 \sin \theta \cdot t - \frac{1}{2} g t^2\). Here, \(y\) is the vertical position, \(v_0 \sin \theta\) gives the initial vertical speed, \(g\) is the acceleration due to gravity (\(9.81 \text{ m/s}^2\)), and \(t\) is time.
The aim is to find a \(t\) where \(y\) matches the final basket height of \(3.05 \text{ m}\) from the initial position of \(2.10 \text{ m}\). Solving for \(t\) allows us to determine how long the ball is in the air, essential for linking back to the horizontal motion equations.

Quadratic Equation in Physics

Quadratic equations frequently appear in physics, especially in projectile motion problems, due to the parabolic path of projectiles.
In our basketball exercise, the vertical motion equation transforms into a quadratic form when substituting known values. For example: \(3.05 = 2.10 + (v_0 \sin 38.0^\circ) \cdot t - \frac{1}{2} \cdot 9.81 \cdot t^2\). Further transformation leads to: \(0.95 = (v_0 \sin 38.0^\circ) \cdot t - \frac{1}{2} \cdot 9.81 \cdot t^2\).
To solve for \(t\), we use the quadratic formula: \[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] where \(a\), \(b\), and \(c\) are coefficients from the quadratic equation derived.Understanding how to solve these equations is vital for calculating the time a projectile spends airborne, providing insight into the correct initial conditions needed for a successful shot.