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 golf ball has an initial position \(\left\langle x_{0}, y_{0}\right\rangle=\langle 0,0\rangle\) when it is hit at an angle of \(30^{\circ}\) with an initial speed of \(150 \mathrm{ft} / \mathrm{s}\).

Short answer

To find the time of flight, set y=0 in the vertical position equation: \(0=75t-16t^2\) Solving for \(t\) gives two roots, \(t=0\) and \(t=T\), where \(T\) corresponds to the time of flight. Range of the object, \(R\), is the horizontal distance traveled during the time of flight: \(R=x_0+v_{x0}T\))

Step-by-step solution

(Step 1: Initialize the given data)

(We know the initial position vector, \(\left\langle x_{0}, y_{0}\right\rangle=\langle 0,0\rangle\), the initial speed \(v_0=150 \mathrm{~ft/s}\), and the angle of projection, \(\theta=30^{\circ}\). Use these values for our calculations.)

(Step 2: Find the initial velocity components)

(Determine the components of the initial velocity vector, separating it into horizontal (\(v_{x0}\)) and vertical (\(v_{y0}\)) components: \(v_{x0}=v_{0}\cos \left(\theta\right)=150\cos(30^{\circ})=150\left(\frac{\sqrt{3}}{2}\right)=75\sqrt{3}\mathrm{~ft/s}\) \(v_{y0}=v_{0}\sin \left(\theta\right)=150\sin(30^{\circ})=150\left(\frac{1}{2}\right)=75\mathrm{~ft/s}\))

(Step 3: Find the velocity and position vectors)

(From the principles of projectile motion: Velocity vector components: \(v_x=v_{x0}=75\sqrt{3}\text{ ft/s}\) \(v_y=v_{y0}-gt=75-32t\text{ ft/s}\) Position vector components: \(x=x_0+v_{x0}t=0+75\sqrt{3}t\text{ ft}\) \(y=y_0+v_{y0}t-\frac{1}{2}gt^2=0+75t-16t^2\text{ ft})\)

(Step 4: Graph the trajectory)

(To graph the trajectory, eliminate \(t\) between position equations \(x\) and \(y\). Solve \(t\) in terms of \(x\) from the horizontal position equation and substitute it in the vertical position equation: \(t=\frac{x}{75\sqrt{3}}\) \(y=75\frac{x}{75\sqrt{3}}-16\left(\frac{x}{75\sqrt{3}}\right)^2 = \frac{x}{\sqrt{3}}-\frac{8}{225}x^2\) The graph of this equation represents the trajectory.)

(Step 5: Time of flight and range of the object)

(Time of flight occurs when the object reaches the ground, i.e., \(y=0\): \(0=\frac{x}{\sqrt{3}}-\frac{8}{225}x^2\) To find the time of flight, set \(y=0\) in the vertical position equation: \(0=75t-16t^2\) Solving for \(t\) gives two roots, \(t=0\) and \(t=T\), where \(T\) corresponds to the time of flight. Range of the object, \(R\), is the horizontal distance traveled during the time of flight: \(R=x_0+v_{x0}T\))

(Step 6: Maximum height of the object)

(The maximum height of the object is reached when the vertical velocity component becomes 0: \(0=v_{y0}-gT_h\) Solve for \(T_h\), where \(T_h\) represents the time when maximum height is reached. The maximum height, \(H_{max}\), can be obtained by substituting \(T_h\) into the vertical position equation: \(H_{max} = 0+75T_h-16T_h^2\))

Key concepts

Velocity and Position Vectors

Understanding the velocity and position vectors of an object in projectile motion is crucial for predicting its future location at any given time. These vectors represent the object's speed and direction (velocity) and its coordinates in space (position) as functions of time.

For our golf ball, with an initial speed of 150 ft/s at a 30° angle, the initial velocity components are calculated using trigonometry: horizontally, it's \(v_{x0} = 150\cos(30^{\textdegree}) = 75\sqrt{3} \text{ ft/s}\); vertically, it's \(v_{y0} = 150\sin(30^{\textdegree}) = 75 \text{ ft/s}\). This separation into x and y components is especially helpful in projectile motion because the object moves independently in these directions due to gravity only affecting the vertical component.

The position vectors change over time. Horizontally, the distance increases uniformly since the horizontal velocity \(v_{x0}\) remains constant, given by \(x = 75\sqrt{3}t \text{ ft}\). Vertically, the position changes with time due to gravity's influence, given by \(y = 75t - 16t^2 \text{ ft}\). This vertical motion results in a parabolic trajectory.

Graphing Trajectories

Graphing the trajectory of a projectile offers a visual representation of its path through space. It helps in understanding how the horizontal and vertical motions combine to create the object's parabolic flight.

To graph the trajectory, we eliminate time (\(t\)) from the position equations to find a relationship between \(x\) and \(y\). We get a quadratic equation, \(y = \frac{x}{\sqrt{3}} - \frac{8}{225}x^2\), which is the classic equation of a parabola opening downwards.

By plotting this equation, we can see that the golf ball's path starts at the origin, rises to a peak, and then falls back to ground level. The highest point of the graph corresponds to the maximum height the golf ball reaches in its flight. The trajectory's shape and the distance traveled depend on the initial velocity components and gravity.

Time of Flight

The time of flight for a projectile is the total time it spends in the air before it lands back on the ground. Determining this time is critical for predicting how far an object will travel horizontally during its journey.

For the golf ball, to find the time of flight, we rely on the vertical position formula, initially setting it to zero (when the ball hits the ground again): \(0 = 75t - 16t^2\). The non-zero solution to this equation gives us the total time the ball remains in the air. Next, to find the range of the golf ball, we use the time of flight in the horizontal position equation, which tells us how far the golf ball travels before it lands.

Maximum Height of Projectile

The maximum height of a projectile is the highest vertical distance it reaches relative to its starting point. This occurs when the vertical component of the projectile's velocity is zero, momentarily pausing before beginning its descent.

To calculate the maximum height that our golf ball achieves, we first find the time (\(T_h\)) at which the vertical velocity \(v_y\) becomes zero: \(0 = 75 - 32T_h\). By solving for \(T_h\), we have the exact moment when the ball is at its peak height. Subsequently, we substitute this time into the vertical position equation to find the maximum height, \(H_{max} = 75T_h - 16T_h^2\), representing the apex of the golf ball's path.