Question

An object is fired vertically upward with an initial velocity \(v(0)=v_{0}\) from an initial position \(s(0)=s_{0}\). a. For the following values of \(v_{0}\) and \(s_{0},\) find the position and velocity functions for all times at which the object is above the ground. b. Find the time at which the highest point of the trajectory is reached and the height of the object at that time. $$v_{0}=29.4 \mathrm{m} / \mathrm{s}, s_{0}=30 \mathrm{m}$$

Short answer

When does the object reach the highest point, and what is its height at that point? Answer: The position function is given by \(s(t) = 30 + 29.4t - 4.905t^2\), and the velocity function is given by \(v(t) = 29.4 - 9.81t\). The object reaches its highest point at approximately \(t = 3.00\) seconds, and its height at that point is approximately \(74.1\) meters above the ground.

Step-by-step solution

Establish the equations of motion for the problem

The equations of motion we will use for this problem are: 1. \(v(t) = v_0 - gt\), where \(v(t)\) is the velocity at time \(t\), \(v_0\) is the initial velocity, and \(g\) is the acceleration due to gravity. We use a negative sign for \(g\) since the acceleration is acting in the downward direction. 2. \(s(t) = s_0 + v_0 t - \frac{1}{2}gt^2\), where \(s(t)\) is the position at time \(t\), \(s_0\) is the initial position, and \(g\) is the acceleration due to gravity. In this problem, we have \(v_0 = 29.4\) m/s, \(s_0 = 30\) m, and \(g = 9.81\) m/s².

Calculate the velocity function

Using the first equation of motion, we find the velocity function: \(v(t) = v_0 - gt = 29.4 - 9.81t\)

Calculate the position function

Using the second equation of motion, we find the position function: \(s(t) = s_0 + v_0t - \frac{1}{2}gt^2 = 30 + 29.4t - 4.905t^2\)

Find the time at which the highest point is reached

The highest point is reached when the velocity of the object becomes zero. So, we set the velocity function equal to zero and solve for \(t\): \(0 = 29.4 - 9.81t\) \(t = \frac{29.4}{9.81} \approx 3.00\) s So, the highest point is reached at \(t \approx 3.00\) s.

Find the height of the object at the highest point

To find the height of the object at the highest point, we plug the time \(t \approx 3.00\) s into the position function: \(s(3) = 30 + 29.4(3) - 4.905(3)^2 \approx 74.1\) m The height of the object at the highest point is approximately \(74.1\) m above the ground.

Key concepts

Equations of Motion

In kinematics, equations of motion describe how an object's position and velocity change over time. They are essential in understanding the movement of objects under uniform acceleration, such as projectiles. Let's explore the equations that help us determine both position and velocity:

• Velocity Equation: The velocity of an object as time progresses can be expressed as \(v(t) = v_0 - gt\). Here, \(v_0\) is the initial velocity, \(g\) is the acceleration due to gravity, and \(t\) is time.
• Position Equation: The position of the object as it moves can be determined using \(s(t) = s_0 + v_0t - \frac{1}{2}gt^2\). \(s_0\) is the initial position of the object.

By using these equations, one can predict future positions and velocities of the object through variations in time, which is crucial for solving problems related to motion.

Initial Velocity

Initial velocity, denoted by \(v_0\), is the speed at which an object begins its motion. In this context, it's their velocity at time \(t=0\). Understanding initial velocity is vital because:

• It sets the stage for future calculations, such as how long the object will be in motion and how high it will rise.
• In our specific example, the object starts with an initial velocity \(v_0 = 29.4\, \text{m/s}\), meaning it was released with a specific force, complying with this speed.

Initial velocity is a crucial factor that affects the object's overall trajectory and eventual motion behavior.

Acceleration Due to Gravity

The constant pull exerted by the Earth's gravity on all objects is known as the acceleration due to gravity, denoted as \(g\). For most problems on Earth's surface, \(g\) is standardized to \(9.81\, \text{m/s}^2\), directed downward.

• This value plays a significant role in influencing how fast an object accelerates or decelerates when in vertical motion.
• Since it is a downward force, it opposes any upward velocity, continuously reducing it until the object momentarily stops before descending.

Recognizing the effect of gravity is crucial as it is an ever-present force, dictating the object's motion pattern in kinematics.

Maximum Height Calculation

Finding the maximum height is key to understanding the entire trajectory of the object. The maximum point is reached when the object's instantaneous velocity becomes zero, due to the consistent downward pull of gravity. To find this point:

• Set the velocity equation to zero. In our case, \(v(t) = 29.4 - 9.81t = 0\) helps us solve for \(t\).
• This results in \(t = 3.00\, \text{s}\), indicating the time when the object reaches the peak of its path.
• Substitute \(t = 3.00\) into the position function, \(s(t) = 30 + 29.4t - 4.905t^2\), to find the maximum height, which computes to \(s(3) \approx 74.1 \text{m}\).

Understanding maximum height is essential for comprehending the full dynamics of an object's motion, particularly in projectile contexts.