Question

Two-dimensional motion 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 baseball is thrown horizontally from a height of 10 ft above the ground with a speed of \(132 \mathrm{ft} / \mathrm{s}\).

Short answer

Question: A baseball is thrown horizontally from a height of 10 ft above the ground with a speed of 132 ft/s. Determine the velocity and position vectors, graph the trajectory, time of flight and range, and maximum height. Answer: The velocity vector is (132 ft/s) i + (0 ft/s) j. The position vector is (132t) i + (10 - 16.1t²) j. The trajectory can be graphed by plotting the parametric equations x(t) and y(t), forming a parabolic shape. The time of flight is approximately 0.789 s, and the range is about 104.14 ft. The maximum height is 10 ft.

Step-by-step solution

Finding the initial velocity vector and separating it into components.

The initial velocity of the baseball is given as 132 ft/s, and it's thrown horizontally. Therefore, the initial velocity vector can be written as: v_initial = (132 ft/s) i where i is the unit vector in the horizontal direction. We don't have any initial velocity in the vertical direction (j) because the ball is thrown horizontally. So, the initial velocity vector can be also written as: v_initial = (132 ft/s) i + (0 ft/s) j

Calculating position vectors.

We can determine the position vector by integrating the velocity vector with respect to time separately for x and y components: For the x-component: x(t) = x_initial + v_x_initial*t x(t) = 0 + (132 ft/s) * t For the y-component, we need to consider the acceleration due to gravity, which is -32.2 ft/s² (downward): y(t) = y_initial + v_y_initial*t - (1/2)*g*t² y(t) = 10 + 0*t - (1/2)*(-32.2 ft/s²) * t² The position vector at any time t is given by the sum of these components in the i and j directions: r(t) = (132t) i + (10 - 16.1t²) j

Graphing the trajectory.

In this step, you plot the parametric equations x(t) and y(t) on graph paper or use a software to plot them. The graph will show the parabolic trajectory of the baseball.

Time of flight and range.

To find the time of flight, we need to determine when the baseball hits the ground (y(t) = 0): 0 = 10 - 16.1t² t² = 10/16.1 t = √(10/16.1) ≈ 0.789 s Now, we can find the range by plugging the time of flight into the x(t) equation: Range = x(t) = 132 * 0.789 ≈ 104.14 ft

Maximum height.

The maximum height is reached when the vertical velocity component is zero (v_y(t) = 0). We know that: v_y(t) = v_y_initial - g*t 0 = 0 - (-32.2 ft/s²) * t t = 0 (since there's no vertical velocity component at the beginning). So, the maximum height is reached at the initial moment, which is given by the initial height above the ground (y_initial): Maximum height = 10 ft To summarize: a) Velocity vector: (132 ft/s) i + (0 ft/s) j, Position vector: (132t) i + (10 - 16.1t²) j b) Graph the parametric equations x(t) and y(t) to show the parabolic trajectory. c) Time of flight: 0.789 s, Range: 104.14 ft d) Maximum height: 10 ft

Key concepts

Velocity and Position Vectors

Understanding the concepts of velocity and position vectors is fundamental in analyzing the motion of objects in physics. In the context of two-dimensional motion, the velocity vector represents both the speed and direction of an object's movement, while the position vector represents the location of an object in relation to a fixed point of origin over time.

For instance, if a baseball is thrown with a known velocity horizontally, its initial velocity vector is fully represented in the horizontal component with no vertical component since there is no initial vertical motion. Mathematically, this is denoted as \(v_{\text{initial}} = (132 \, \text{ft/s}) \mathbf{i} + (0 \, \text{ft/s}) \mathbf{j}\).

The position vector tracks the baseball's location at any given time. It is obtained by integrating the velocity vector. Considering gravity, the vertical position changes by \(\frac{1}{2}gt^2\) where \(g\) is the acceleration due to gravity. Thus, we get the position vector expressed as \(\mathbf{r}(t) = (132t) \mathbf{i} + (10 - 16.1t^2) \mathbf{j}\), representing the horizontal and vertical displacement over time, respectively.

Trajectory Graphing

The trajectory graphing of an object in two-dimensional motion is a visual representation of its path through space. This graphical representation helps to visualize the behavior of the object's motion, which is subject to the forces acting upon it, such as gravity. For the thrown baseball, its trajectory is parabolic due to the influence of gravity pulling it downward.

By plotting the parametric equations \(x(t)\) and \(y(t)\), a clear picture of the movement is formed. This graph makes evident the parabolic nature of the baseball's path from the point of release until it hits the ground. This visual aid is not only valuable for understanding the actual flight path but is also instrumental for further analysis of the motion, such as finding the range or maximum height.

Time of Flight Calculation

Determining the time of flight calculation of a projectile is crucial for understanding when the object will land. This is calculated by setting the vertical position equation equal to zero, which corresponds to the moment when the object touches ground level again.

For the baseball, we solve the equation \(0 = 10 - 16.1t^2\) to find the time when the baseball hits the ground. Upon solving, we find that the time of flight is approximately \(0.789 \, \text{s}\). This time of flight, along with the horizontal velocity, can then be used to determine the range or horizontal distance traveled by the baseball, which, in this case, is \(104.14 \, \text{ft}\). These calculations are essential for predicting the full horizontal distance an object will cover while it is airborne.

Maximum Height Determination

The maximum height determination is a concept used to ascertain the highest vertical point reached by a projectile. In scenarios where an object is projected upwards, the highest point is attained when the vertical velocity equals zero, which is when the gravitational pull negates the initial vertical component of the velocity. However, in cases where the initial vertical velocity is zero, such as the horizontally thrown baseball, it does not ascend beyond its initial release height.

In this exercise, where the baseball is thrown horizontally from a certain height, the maximum height corresponds to the initial height, which is \(10 \, \text{ft}\). If an object was thrown with an initial vertical velocity, the calculation would involve finding the time it takes for gravity to decelerate the object to a stop vertically and then substituting this time back into the vertical position equation to find the height.