Question

Consider the motion of the following objects. Assume the \(x\) -axis is horizontal, the positive y-axis is vertical, the ground is horizontal, and only the gravitational force acts on the object. a. Find the velocity and position vectors, for \(t \geq 0\) b. Graph the trajectory. c. Determine the time of flight and range of the object. d. Determine the maximum height of the object. A bascball has an initial position (in feet) of \(\left\langle x_{0}, y_{0}\right\rangle=\langle 0,6\rangle\) when it is thrown with an initial velocity of \(\left\langle u_{0}, v_{0}\right\rangle=\langle 80,10\rangle \mathrm{ft} / \mathrm{s}\)

Short answer

Answer: The parametric equations of motion for the given initial conditions are \(x(t) = 80t\) and \(y(t) = 6 + 10t - 16t^2\).

Step-by-step solution

Setup the equations of motion

The acceleration due to gravity is constant, acting downward with magnitude \(g=32\,\mathrm{ft/s^2}\). So we have that: \(\vec{a}(t)=\langle0,-g\rangle=\langle0,-32\rangle\) ft/s² Now, we need to set up the equations for velocity and position vectors using these equations of motion: \(\vec{v}(t)=\vec{v_0}+\int{\vec{a}(t) \, dt}\) \(\vec{r}(t)=\vec{r_0}+\int{\vec{v}(t) \, dt}\)

Find the velocity vector

From the given initial values \(\left\langle u_{0}, v_{0}\right\rangle=\langle 80,10\rangle \mathrm{ft} / \mathrm{s}\), we can substitute the acceleration to find the velocity vector: \(\vec{v}(t)=\langle 80,10\rangle+\langle0,-32\rangle\cdot t=\langle 80,10-32t \rangle \mathrm{ft} / \mathrm{s}\)

Find the position vector

Using the velocity vector we just found, we can integrate to find the position vector: \(\vec{r}(t)=\langle 0,6\rangle+\int{\langle 80,10-32t \rangle \, dt}=\langle 80t,6+10t-16t^2 \rangle\,\mathrm{ft}\)

Parametric equations for \(x\) and \(y\)

We can rewrite the position vector as parametric equations of motion: \(x(t) = 80t\) \(y(t) = 6 + 10t - 16t^2\)

Find the time of flight

The time of flight is the time when the object hits the ground (i.e., when \(y(t) = 0\)). So we solve the equation \(y(t) = 0\) for \(t\): \(6+10t-16t^2=0\) Solve the quadratic equation for \(t\) to get two roots: \(t_1\) and \(t_2\). The time of flight is the positive root.

Find the range

The range is the horizontal distance the baseball travels before hitting the ground. To find the range, simply substitute the time of flight \(t\) from step 5 into the equation for \(x(t)\): \(Range = 80\cdot t\)

Find the maximum height

To find the maximum height, we can consider that the vertical velocity component is zero at the peak. So we find the time \(t\) when \(v_y(t)=0\) (i.e., \(10-32t=0\)) and then substitute this time into the equation for \(y(t)\).

Graph the trajectory

To graph the trajectory, we can use a graphing tool to plot the parametric equations \(x(t)=80t\) and \(y(t)=6+10t-16t^2\), and include the range and maximum height found above. Remember that the graph should only display for \(t\geq0\). After completing these steps, you should have a complete analysis of the baseball's motion, including the velocity and position vectors, trajectory, time of flight, range, and maximum height.

Key concepts

Velocity Vector Projectile

The velocity vector of a projectile reveals how its speed and direction change over time. In our baseball example, its initial velocity is given by the vector \<\langle u_{0}, v_{0}\rangle=\langle 80,10\rangle\> ft/s. Gravity's effect on the vertical component is constant and downwards, represented by the acceleration vector \<\vec{a}(t)=\langle0,-32\rangle\> ft/s². Thus, the velocity vector at any time \(t\) is calculated by adding the product of acceleration and time to the initial velocity, yielding \<\vec{v}(t)=\langle 80,10-32t \rangle\> ft/s. This expression tells us that while the horizontal velocity remains constant at 80 ft/s, the vertical velocity decreases by 32 ft/s every second due to gravity.

Position Vector Projectile

The position vector of a projectile describes its location at any given moment. For the projectile in our problem, the initial position is \<\langle x_{0}, y_{0}\rangle=\langle 0,6\rangle\> feet. To find the position at any time \(t\), we integrate the velocity vector which results in the position vector \<\vec{r}(t)=\langle 80t,6+10t-16t^2 \rangle\> feet. This equation conveys that the projectile moves linearly in the horizontal direction and follows a parabolic trajectory in the vertical direction due to the influence of gravity.

Time of Flight

The time of flight for a projectile is the duration it remains airborne before returning to ground level. For our baseball, the ground level corresponds to a vertical position \(y(t)\) of 0. By solving the quadratic equation \(6+10t-16t^2=0\), we determine the time at which the baseball reaches ground level again. Only the positive root of this equation represents a physical solution, giving us the time of flight. This value is crucial in calculating other characteristics such as range and maximum height.

Range of Projectile

The range of the projectile is the total horizontal distance covered during its flight. Upon deriving the time of flight, we substitute this value into the horizontal component of the position vector to get the range, represented by the equation \(Range=80\cdot t\). Therefore, the range is directly proportional to the time the baseball spends in the air multiplied by its constant horizontal velocity.

Maximum Height Projectile

A projectile's maximum height is the greatest vertical distance it reaches. To find this for our baseball, we need to identify when the vertical component of the velocity vector becomes zero, which happens at the peak of the trajectory. Solving \(10-32t=0\) gives us the time at which the maximum height occurs. Substituting this time back into the vertical component of the position vector gives us the maximum height the baseball achieves.

Parametric Equations Projectile

Parametric equations express the coordinates of the projectile as functions of time. They give us a precise mathematical description of the path the projectile takes through space. Our problem provides us with two parametric equations: \(x(t)=80t\) for the horizontal movement, and \(y(t)=6+10t-16t^2\) for the vertical movement. These equations highlight the independence of horizontal and vertical motions, and when combined, describe the trajectory of the projectile.

Quadratic Equation Projectile

Quadratic equations are crucial in projectile motion as they describe the vertical displacement of a projectile affected by gravity. In our scenario, the quadratic equation \(y(t) = 6 + 10t - 16t^2\) is used to calculate important parameters like time of flight and maximum height. By solving this equation, we can also plot the trajectory of the baseball as it moves under the influence of gravity, revealing a parabolic path.