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. $$v_{0}=49 \mathrm{m} / \mathrm{s}, s_{0}=60 \mathrm{m}$$

Short answer

Short Answer: To find the height of the object at its highest point, you must substitute the time value t = 49/9.8 into the position function s(t): $$s\left(\frac{49}{9.8}\right)=-4.9\left(\frac{49}{9.8}\right)^2+49\left(\frac{49}{9.8}\right)+60$$ Solve this equation to obtain the maximum height of the object.

Step-by-step solution

Define the Position and Velocity Functions

Considering the gravitational force, we can define the acceleration function as $$a(t)=-g$$ where \(g\) represents the acceleration due to gravity (\(9.8m/s^2\)). Now, we can determine the velocity function by integrating the acceleration function: $$v(t)=\int(-g)dt=-gt+C_{1}$$ Since \(v(0) = v_{0}\), we can find the constant \(C_{1}\): $$-g(0)+C_{1}=v_{0} \Rightarrow C_{1}=v_{0}$$ So the velocity function is: $$v(t)=-gt+v_{0}$$ Next, we define the position function by integrating the velocity function: $$s(t)=\int(-gt+v_{0})dt=-\frac{1}{2}gt^2+v_{0}t+C_{2}$$ Given that \(s(0)=s_{0}\), we determine the constant \(C_{2}\): $$-\frac{1}{2}g(0)^2+v_{0}(0)+C_{2}=s_{0} \Rightarrow C_{2}=s_{0}$$ The resulting position function is: $$s(t)=-\frac{1}{2}gt^2+v_{0}t+s_{0}$$

Find the Position and Velocity Functions for the Given Initial Conditions

Given the initial conditions \(v_{0}=49m/s\) and \(s_{0}=60m\), we can write the position and velocity functions as: $$v(t)=-9.8t+49$$ $$s(t)=-4.9t^2+49t+60$$

Determine the Time at Which the Highest Point Is Reached

The highest point of the trajectory is reached when the velocity becomes zero. Thus, we need to find when \(v(t)=0\): $$0=-9.8t+49$$ To find the time \(t\), solve the equation: $$t=\frac{49}{9.8}$$

Calculate the Height of the Object at the Highest Point

Now that we have the time at which the highest point is reached, we can find the height of the object at that time using the position function: $$s\left(\frac{49}{9.8}\right)=-4.9\left(\frac{49}{9.8}\right)^2+49\left(\frac{49}{9.8}\right)+60$$ Solve the equation to find the height of the object at the highest point.

Results

1) The position and velocity functions for the given initial conditions are: $$v(t)=-9.8t+49$$ $$s(t)=-4.9t^2+49t+60$$ 2) The time at which the highest point of the trajectory is reached is: $$t=\frac{49}{9.8}$$ 3) The height of the object at the highest point can be found by solving: $$s\left(\frac{49}{9.8}\right)$$

Key concepts

Velocity Function

Understanding the velocity function is crucial when studying motion in a gravitational field. The velocity function, denoted as \( v(t) \), describes how the velocity of an object changes over time. It's derived by integrating the acceleration due to gravity, which is constant at \(-g\), with respect to time. As seen in the given solution, initially, the object has an upward velocity of \( v_0 \), and as time passes, the gravitational pull decreases its upward velocity until it eventually starts descending.
For an object thrown vertically upward with an initial velocity \( v_0 \), the velocity function is expressed as \( v(t) = -gt + v_0 \). This equation factors in the constant downward pull of gravity, represented by \(-g\), as it works against the object's initial upward velocity. The outcome of this function allows us to predict the velocity at any given time after the object is propelled.

Position Function

The position function, represented by \( s(t) \), describes the object's position over time while it moves under the influence of gravity. By integrating the velocity function, we obtain the position function, which expresses the object's displacement from its initial position \( s_0 \). The step-by-step solution shows the integration process, leading to the formula \( s(t) = -\frac{1}{2}gt^2 + v_0t + s_0 \).
This kinematic equation considers the initial velocity \( v_0 \) and initial position \( s_0 \), and it incorporates the constant acceleration due to gravity to determine how far from its starting point the object will be at any moment in time. Importantly, the negative sign in front of the gravity term reflects the deceleration effect of gravity on the object's ascent and its acceleration effect during the object's fall back to the ground.

Acceleration Due to Gravity

Acceleration due to gravity, commonly symbolized as \( g \), is a constant that defines the rate at which an object accelerates towards the Earth when the only force applied is gravity. On the surface of the Earth, this value is approximately 9.8 meters per second squared (\( 9.8 m/s^2 \)).
This acceleration is a key component in the kinematic equations of motion and is considered to be constant for all objects near the Earth's surface, irrespective of their mass. In the solution for our exercise, the negative sign in the term \(-g\) indicates that gravity is acting in the opposite direction to the initial motion of the object when it is propelled upward.

Kinematics Equations

Kinematics equations are tools that allow us to describe the motion of objects using mathematical formulas. They comprise formulas for velocity, acceleration, displacement, and time, connecting these variables in ways that can be applied to predict future motion given certain initial conditions.
In the context of vertical motion in a gravitational field, we primarily work with two kinematic equations: the velocity function \( v(t) = -gt + v_0 \) and the position function \( s(t) = -\frac{1}{2}gt^2 + v_0t + s_0 \). These equations represent the linear relationship between time and velocity, and the quadratic relationship between time and position, respectively. They are fundamental in solving problems involving objects moving under the influence of gravity, as seen in our exercise.

Trajectory Optimization

Trajectory optimization is concerned with finding the best path or trajectory for an object to reach a certain point in space, or to achieve a certain objective, such as maximum height or range. It involves using kinematic equations and calculus to maximize or minimize specific parameters of motion.
In our given problem, the optimization aspect is to find the highest point of the trajectory, which corresponds to the maximum value of the position function \( s(t) \). As the solution outlines, this maximum is found when the velocity function reaches zero \( v(t) = 0 \), signaling that the object has reached its peak and will start to fall back. The time at which this occurs, as well as the height at this maximum point, are the optimized values sought in this scenario. Using trajectory optimization principles, one can also solve for other conditions, such as the least amount of energy expended or the shortest time taken to reach a certain height.