Problem 9 - College Physics

Question

A tennis ball rolls off the edge of a tabletop 0.750 m above he floor and strikes the floor at a point 1.40 m horizontally from the edge of the table. (a) Find the time of flight of the ball. (b) Find the magnitude of the initial velocity of the ball.(c) Find the magnitude and direction of the velocity of the ball just before it strikes the floor.

Step-by-Step Solution

Answer

(a) 0.391 s. (b) 3.58 m/s. (c) 5.23 m/s at 46.5° below horizontal.

Step 1: Analyzing the Vertical Motion

First, analyze the vertical motion of the ball since it's affected by gravity. Use the equation for vertical displacement, \( y = v_{i_y} t + \frac{1}{2}gt^2 \). Given that the ball rolls off horizontally, \( v_{i_y} = 0 \). We have \( y = -0.750 \) m and \( g = 9.81 \) m/s². Simplifying, we have \( -0.750 = \frac{1}{2}(9.81)t^2 \). We can solve for \( t \).

Step 2: Solving for Time of Flight

Solving the equation \( -0.750 = \frac{1}{2}(9.81)t^2 \) yields \( t^2 = \frac{-2 \times -0.750}{9.81} \). Calculate \( t^2 \) and then take the square root to find \( t \). The result is \( t \approx 0.391 \) s.

Step 3: Analyzing the Horizontal Motion

Use the horizontal displacement equation, \( x = v_{i_x}t \), where \( x = 1.40 \) m and \( t = 0.391 \) s. Solve for \( v_{i_x} \) as \( v_{i_x} = \frac{x}{t} \).

Step 4: Calculating Initial Velocity

Solve for the initial velocity using \( v_{i_x} = \frac{1.40}{0.391} \), resulting in \( v_{i_x} \approx 3.58 \) m/s. Since the ball rolls off horizontally, \( v_i = v_{i_x} = 3.58 \) m/s.

Step 5: Calculating Final Vertical Velocity

The vertical component of the velocity just before the ball hits the ground can be found using \( v_{f_y} = v_{i_y} + gt \). With \( v_{i_y} = 0 \), \( v_{f_y} = 9.81 \times 0.391 \approx 3.84 \) m/s.

Step 6: Calculating Magnitude of Final Velocity

The magnitude of the velocity before impact, \( v_f \), is given by \( v_f = \sqrt{v_{i_x}^2 + v_{f_y}^2} \). Substitute \( v_{i_x} = 3.58 \) m/s and \( v_{f_y} = 3.84 \) m/s to find \( v_f \approx 5.23 \) m/s.

Step 7: Calculating Direction of Final Velocity

To find the direction, compute the angle \( \theta \) using \( \tan(\theta) = \frac{v_{f_y}}{v_{i_x}} \). With \( v_{f_y} = 3.84 \) m/s and \( v_{i_x} = 3.58 \) m/s, \( \theta = \tan^{-1}\left(\frac{3.84}{3.58}\right) \approx 46.5^\circ \) below the horizontal.

Key Concepts

Vertical Motion

When analyzing projectile motion, it's crucial to separate vertical and horizontal motions.
Vertical motion is influenced by gravity, which acts downward. In the case of the tennis ball rolling off the table, the vertical motion starts with no initial vertical velocity.

• The initial vertical velocity, denoted as \( v_{i_y} \), is zero because the ball rolls off horizontally.
• The vertical displacement \( y \) is the height of the table, which is -0.750 meters (the negative sign indicates motion downward).
• Gravity \( g \) accelerates the ball at 9.81 m/s² downward.

The vertical motion equation, \( y = v_{i_y} t + \frac{1}{2}gt^2 \), simplifies to \( -0.750 = \frac{1}{2}(9.81)t^2 \).
Solving this equation helps find the time of flight, a key aspect of understanding how long the ball takes to hit the floor.

Horizontal Motion

Horizontal motion in projectile scenarios is typically unaffected by acceleration due to gravity. In the exercise, since the ball rolls off horizontally, the initial horizontal velocity \( v_{i_x} \) remains constant throughout its flight.

• Horizontal displacement \( x \) is simply how far the ball lands from the table's edge; here, it's 1.40 meters.
• The initial horizontal velocity can be determined using the formula \( x = v_{i_x} t \), where \( t \) is the time of flight found using the vertical motion equations.

Calculating \( v_{i_x} = \frac{x}{t} \) yields the ball's horizontal speed, further allowing the assessment of the ball's motion across its path. The consistent horizontal speed is key in combinations with vertical motion to understand the ball's trajectory.

Initial Velocity

Initial velocity in projectile motion describes the speed and direction an object has right as it begins its flight. For this scenario with the tennis ball, we concentrate on the horizontal component because it rolls off the table horizontally.

• Since the ball does not initially move vertically, its initial vertical component \( v_{i_y} \) is zero.
• The calculated horizontal component \( v_{i_x} = 3.58 \) m/s serves as its initial velocity, owing to the absence of initial vertical motion.
• This means the initial velocity is purely horizontal, allowing easy calculation for angles and speeds later as part of combined velocity components when the ball is in flight.

Understanding the initial velocity is fundamental, as it characterizes the launch conditions, setting the stage for calculating other motion characteristics as the ball travels.

Final Velocity Direction

Once a projectile like a tennis ball nears the end of its flight, its velocity has both vertical and horizontal components due to gravity. The final velocity direction before it hits the floor is determined by analyzing both components:

• The final vertical velocity \( v_{f_y} \) is calculated as \( v_{i_y} + gt = 3.84 \) m/s, since \( v_{i_y} = 0 \) initially.
• The final horizontal velocity \( v_{i_x} = 3.58 \) m/s remains unchanged, as horizontal velocity stays constant in projectile motion.
• The magnitude of the final velocity \( v_f \) is \( \sqrt{v_{i_x}^2 + v_{f_y}^2} \approx 5.23 \) m/s.
• The direction, or angle \( \theta \), is found using \( \tan(\theta) = \frac{v_{f_y}}{v_{i_x}} \), resulting in approximately \( 46.5^\circ \) below the horizontal.

Understanding the final velocity direction helps in visualizing the motion path the ball takes before impact, combining both gravity's pull and the initial horizontal motion mechanics.