Question

Motion in a gravitational field 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 \((s=0).\) 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

Answer: The highest point reached by the object is approximately 74.1 m and it takes about 3.00 seconds to reach this point.

Step-by-step solution

Position and Velocity Functions

The object is experiencing a downward acceleration due to gravity, which we represent as \(g\). Generally, \(g\) is taken as approximately \(9.8 \ \text{m/s}^2\). As it is acting downward we will use the negative sign, i.e., \(-9.8 \ \text{m/s}^2\). The general equation for an object undergoing constant acceleration is given as: $$s(t) = s_0 + v_0t + \frac{1}{2}at^2$$ In our case, the acceleration (\(a\)) is equal to the gravitational acceleration (\(-g\)), which is \(-9.8 \ \text{m/s}^2\). Using the given value of \(s_0 = 30 \ \text{m}\) and \(v_0 = 29.4 \ \text{m/s}\), we get: $$s(t) = 30 + 29.4t - \frac{1}{2}(9.8)t^2$$ The velocity function can be obtained by differentiating the position function with respect to time: $$v(t) = \frac{ds}{dt} = 29.4 - 9.8t$$ Now, we will find the time at which the highest point of the trajectory is reached.

Time to Reach the Highest Point

At the highest point of the trajectory, the velocity of the object becomes 0, as it momentarily stops before starting to descend. Thus, we set \(v(t) = 0\) and solve for \(t\): $$0 = 29.4 - 9.8t$$ Solving for \(t\), we get: $$t = \frac{29.4}{9.8}$$ $$t \approx 3.00 \ \text{s}$$ Now, let's find the height of the object at that time.

Height at the Highest Point

We substitute the value of \(t\) we found into the position function \(s(t)\) to find the height at the highest point: $$s(3.00) = 30 + 29.4(3.00) - \frac{1}{2}(9.8)(3.00)^2$$ Calculating this, we get: $$s(3.00) \approx 74.1 \ \text{m}$$ So, the highest point reached by the object is approximately \(74.1 \ \text{m}\) and it takes about \(3.00 \ \text{s}\) to reach this point. The final results are as follows: 1. Position function: \(s(t) = 30 + 29.4t - 4.9t^2\) 2. Velocity function: \(v(t) = 29.4 - 9.8t\) 3. Time to reach the highest point: \(t \approx 3.00 \ \text{s}\) 4. Height at the highest point: \(s(3.00) \approx 74.1 \ \text{m}\)

Key concepts

Gravitational Acceleration

Gravitational acceleration is a fundamental concept in physics, particularly in the study of projectile motion. It refers to the constant acceleration that an object experiences when it is in freefall near the surface of a celestial body, such as Earth. On Earth, the value of gravitational acceleration is approximately \(9.8 \, \text{m/s}^2\), and this acts downwards towards the center of the planet.
When considering projectile motion, gravitational acceleration plays a crucial role in determining both the trajectory and time of flight of an object. It affects how quickly the velocity of the object changes over time. Because gravity acts downward, it is often represented with a negative sign in equations when the object is moving upwards.
Understanding gravitational acceleration is essential when calculating the position and velocity of an object. It allows us to predict how high and how fast an object will travel when launched either vertically, like in this problem, or at an angle. These predictions are crucial not only for solving textbook problems but also in real-world applications like engineering and sports.

Position Function

The position function describes how the position of an object changes over time due to its velocity and acceleration. In projectile motion, this is essential for determining where an object will be at any given time after being launched.

In the problem, the position function is governed by the equation:

• \(s(t) = s_0 + v_0t + \frac{1}{2}at^2\)

Here, \(s_0\) is the initial position, \(v_0\) is the initial velocity, and \(a\) is the acceleration due to gravity, which is \(-9.8 \, \text{m/s}^2\) for objects near Earth's surface.

When we plug the initial conditions \(s_0 = 30 \, \text{m}\) and \(v_0 = 29.4 \, \text{m/s}\), into the equation, it results in:

• \(s(t) = 30 + 29.4t - 4.9t^2\)

This formula enables us to calculate the height of the object at any moment in time. For instance, to find how high the object will be after it stops rising, substitute the time at which it stops in the position function.

Velocity Function

The velocity function is a key tool in understanding how the speed and direction of an object change over time as it moves through a gravitational field. Velocity is particularly important when it comes to determining when an object will stop rising and start to fall.
For an object in projectile motion, the velocity function is derived from the position function. By differentiating the position function with respect to time, we get the velocity function. The equation for our specific problem is:

• \(v(t) = 29.4 - 9.8t\)

In this equation, \(29.4 \, \text{m/s}\) is the initial upward velocity, and \(-9.8 \, \text{m/s}^2\) represents the gravitational pull slowing it down as it ascends.
To find out when the object reaches its highest point, we set the velocity equation to zero, \(v(t) = 0\), meaning the object is momentarily at rest before falling back down. Solving for \(t\) tells us the exact time it takes to reach this maximum height. As calculated, \(t \approx 3.00 \, \text{s}\). Knowing the time it takes to reach the highest point allows us to further analyze the object's trajectory regarding height and time on the ground.

Trajectory Analysis

Trajectory analysis is the study of the path that an object follows through space as a function of time. In the context of projectile motion, this involves analyzing the vertical and sometimes horizontal components of motion to understand the object's path.

The most critical factors in trajectory analysis are:

• Initial velocity and position
• Gravitational force
• Time

For the vertical motion of an object fired upward, its highest trajectory point occurs when its upward speed decreases to zero, determined using the velocity function. From this point, the object will begin to fall back under the influence of gravity.

In our case study, with \(v_0 = 29.4 \, \text{m/s}\) and \(s_0 = 30 \, \text{m}\), the highest point reached is approximately \(74.1 \, \text{m}\), as calculated using the position function at \(t \approx 3.00 \, \text{s}\). By understanding and applying trajectory analysis, students can predict anything from simple vertical jumps to more complex projectiles, such as balls launched at angles.