Problem 45 - Fundamentals of Physics

Question

(a) With what speed must a ball be thrown vertically from ground level to rise to a maximum height of \(50 \mathrm{~m} ?\) (b) How long will it be in the air? (c) Sketch graphs of \(y, v,\) and \(a\) versus \(t\) for the ball. On the first two graphs, indicate the time at which \(50 \mathrm{~m}\) is reached.

Step-by-Step Solution

Answer

(a) 31.3 m/s, (b) 6.4 seconds, (c) Graphs: parabola for y, linear for v, constant for a.

Step 1: Understand the Problem

We need to determine the initial speed needed for a ball to reach 50 m in height, the total time the ball is in motion, and sketch the graphs of its position, velocity, and acceleration versus time.

Step 2: Identify Key Equations

We'll use the following kinematic equations for each part of the problem:1. Maximum height formula: \( h = \frac{v_i^2}{2g} \), where \( h \) is the maximum height, \( v_i \) is initial velocity, and \( g \) is the acceleration due to gravity (~9.8 m/s²).2. Time to reach maximum height: \( t = \frac{v_i}{g} \).3. Total air time: \( T = 2t \).4. Sketching involves integrating velocity to get position and differentiating position to get acceleration.

Step 3: Calculate Initial Speed

Using the maximum height formula \( h = \frac{v_i^2}{2g} \), we set \( h = 50 \) m and solve for \( v_i \):\[ 50 = \frac{v_i^2}{2 \times 9.8} \]\[ v_i^2 = 50 \times 19.6 \]\[ v_i = \sqrt{980} \approx 31.3 \text{ m/s} \]

Step 4: Calculate Time in the Air

First, calculate the time to reach the maximum height using \( t = \frac{v_i}{g} \):\[ t = \frac{31.3}{9.8} \approx 3.2 \text{ seconds} \]Total time in the air is \( T = 2t \):\[ T = 2 \times 3.2 = 6.4 \text{ seconds} \]

Step 5: Sketch Graphs

For the position graph \( y(t) \): It will be a parabola opening downwards reaching 50 m at \( t = 3.2 \) s.For the velocity graph \( v(t) \): It will be a straight line with a slope of \(-9.8\) m/s², crossing zero at \( t = 3.2 \) s and changing direction.For the acceleration graph \( a(t) \): It is constant at \(-9.8\) m/s² throughout the motion.

Key Concepts

Kinematic Equations

Kinematic equations are essential in analyzing projectile motion problems. They relate various physical quantities such as velocity, displacement, acceleration, and time. In this exercise, we tackle a ball thrown vertically upwards by utilizing these equations.

• The formula for maximum height is given by: \[ h = \frac{v_i^2}{2g} \]where \(h\) is the maximum height, \(v_i\) is the initial velocity, and \(g\) is acceleration due to gravity (approximately \(9.8 \text{ m/s}^2\)). This equation helps determine how high the ball will rise.


• To find the time to reach maximum height, we utilize the equation: \[ t = \frac{v_i}{g} \]This calculates the time taken for the velocity to reduce to zero at the peak of the ball's motion.


• The total time the ball stays in the air is computed by doubling the time to reach the maximum height, signifying the symmetry of the motion: \[ T = 2t \]

Understanding these kinematic equations lets us predict the trajectory and timing of a projectile or any object in free fall.

Initial Velocity Calculation

Determining the initial velocity is crucial for finding out how fast the ball must be thrown to achieve the desired height in projectile motion. In the problem presented, we want the ball to reach a height of 50 meters. Using the maximum height formula:\[ 50 = \frac{v_i^2}{2 \times 9.8} \]We solve for \(v_i\).
Breaking it down:1. Multiply both sides by \(19.6\) (which is \(2 \times 9.8\)) to clear the fraction:\[ v_i^2 = 50 \times 19.6 \]2. Calculate the product and take the square root to isolate \(v_i\):\[ v_i^2 = 980 \]\\[ v_i = \sqrt{980} \approx 31.3 \text{ m/s} \]
This calculation shows that the initial speed needs to be approximately 31.3 meters per second to reach a 50 meter height. Knowing this velocity helps in planning the throw effectively.

Time of Flight

The time of flight is the duration for which the ball remains in the air from the moment it is thrown until it returns to the ground. This is a fundamental aspect that provides insights on how long the object will be moving.To calculate the time to reach the peak height of 50 meters, we use the formula:\[ t = \frac{v_i}{g} \]Plug in the initial velocity (\(v_i\)) calculated previously and gravity \(g = 9.8 \text{ m/s}^2\):\[ t = \frac{31.3}{9.8} \approx 3.2 \text{ seconds} \]
Since the time of ascent is equal to the time of descent in symmetric projectile motion, the total time of flight \((T)\) is:\[ T = 2t = 2 \times 3.2 = 6.4 \text{ seconds} \]
This symmetric nature of motion implies that the ball spends equal time going up and coming down, simplifying the calculation of total flight duration.

Graphical Analysis of Motion

Visualizing motion through graphs is a powerful method to comprehend projectile dynamics better. Three graphs are generally used:

• Position vs. Time Graph (\(y(t)\)): This graph maps out the height of the ball over time. For a ball thrown vertically, the curve is a downward-opening parabola. The vertex at 3.2 seconds marks the maximum height of 50 meters. This symmetry indicates the time to ascend equals the time to descend.


• Velocity vs. Time Graph (\(v(t)\)): Here, velocity changes linearly over time, represented as a straight line with a negative slope due to constant acceleration from gravity. The line crosses the time axis at 3.2 seconds, depicting when the velocity is zero at the peak of the ascent.


• Acceleration vs. Time Graph (\(a(t)\)): Constant acceleration due to gravity is shown as a flat line at \(-9.8 \text{ m/s}^2\). This reflects that acceleration is consistent regardless of the ball's position in its flight.

These graphical representations enhance your understanding by providing a clearer picture of each stage of the projectile's journey, clarifying how kinematic equations apply over time.