Question

Consider the motion of the following objects. Assume the \(x\) -axis is horizontal, the positive y-axis is vertical and opposite g. 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 projectile is launched from a platform \(20 \mathrm{ft}\) above the ground at an angle of \(60^{\circ}\) with a speed of \(250 \mathrm{ft} / \mathrm{s}\). Assume the origin is at the base of the platform.

Short answer

#tag_title# Step 2: Compute the position and velocity vector equations as functions of time #tag_content# In order to find the position and velocity vector equations as functions of time, we will use the following kinematic equations: \(\overrightarrow{r}(t) = \overrightarrow{r_0} + \overrightarrow{v_0}t - \frac{1}{2}gt^2\hat{j}\) \(\overrightarrow{v}(t) = \overrightarrow{v_0} - gt\hat{j}\) Where \(\overrightarrow{r_0}\) is the initial position vector, \(\overrightarrow{v_0}\) is the initial velocity vector, \(g\) is the acceleration due to gravity \((32 ft/s^2)\), and \(t\) is time. Substituting the values for the initial position and velocity vectors, we get: \(\overrightarrow{r}(t) = \langle 0, 20\rangle + \langle 125, 125\sqrt{3}\rangle t - \frac{1}{2}(32)t^2\hat{j}\) \(\overrightarrow{v}(t) = \langle 125, 125\sqrt{3}\rangle - 32t\hat{j}\) These equations give the position and velocity of the projectile as functions of time.

Step-by-step solution

Find the initial position and velocity vectors

Given that the projectile is launched from a platform \(20 ft\) above the ground, its initial position is \(\overrightarrow{r_0} = \langle 0, 20\rangle\). The initial speed of the projectile is given as \(250 ft/s\) and the launch angle is \(60^{\circ}\). We can find the initial velocity vectors components as follows: \(v_{0x} = v_0 \cos(60^{\circ}) = 250 \cos(60^{\circ}) = 125 ft/s\) \(v_{0y} = v_0 \sin(60^{\circ}) = 250 \sin(60^{\circ}) = 250 \cdot \frac{\sqrt{3}}{2} = 125\sqrt{3} ft/s\) So, the initial velocity vector is \(\overrightarrow{v_0} = \langle 125, 125\sqrt{3}\rangle\).

Key concepts

Velocity Vectors

Velocity vectors are essential in determining how fast and in what direction a projectile is moving. When a projectile is launched, it has both horizontal and vertical components to its velocity. The horizontal component (\( v_{0x} \)) is responsible for the movement along the x-axis, while the vertical component (\( v_{0y} \)) governs the movement along the y-axis.

These components can be figured out using trigonometry, as they depend on the initial speed and the angle of projection. For a launch speed \( v_0 \) and a launch angle \( \theta \), the equations are:

• Horizontal velocity: \( v_{0x} = v_0 \cos(\theta) \)
• Vertical velocity: \( v_{0y} = v_0 \sin(\theta) \)

These vectors are crucial for predicting the path of the projectile over time, a path heavily influenced by the force of gravity after launch.

Position Vectors

Position vectors give us information about where a projectile is at a specific time. The initial position vector can be considered as the starting point of the projectile. In our context, this depends on the height from which the projectile is launched.

From the exercise, the initial position vector is \( \overrightarrow{r_0} = \langle 0, 20 \rangle \mathtt{ft} \), which indicates it starts at the base horizontally and 20 feet vertically above the ground. As time goes on, the position changes according to:

• Horizontal position: \( x(t) = v_{0x} \cdot t \)
• Vertical position: \( y(t) = v_{0y} \cdot t - \frac{1}{2}gt^2 \)

The negative sign in the vertical equation represents the gravitational pull pulling the projectile downward over time.

Graphing Trajectories

Graphing the trajectory of a projectile helps visualize its path through the air. This path is often parabolic due to the influence of gravity, which accelerates the projectile downward.

When graphing:

• The x-axis typically represents time, while the y-axis represents height or range.
• The curve starts at the initial height and describes an arc upward, then downward.

To create this graph, you calculate position vectors over regular time intervals. Plotting these points gives a trajectory.

This helps us see critical elements such as maximum height and where the projectile will land, which are key points in determining the overall behavior of the projectile's motion.

Time of Flight

The time of flight is the total time that the projectile is in the air. It is imperative to calculate this to understand the full range of the projectile's journey.

The time of flight can be determined from the vertical motion equation of the projectile. Set the final vertical position (when it hits the ground, y = 0) and solve for time. When the projectile lands, it reaches y = 0, thus:

• Final equation: \( 20 + (v_{0y} \cdot t) - \frac{1}{2}g t^2 = 0 \)

Solving this quadratic equation will give you the time of flight.

Understanding the time span from launch to landing not only defines how long the object remains airborne, but it's also essential for determining the maximum range and peak height.

Maximum Height

The maximum height of a projectile occurs when the vertical component of its velocity becomes zero. This is the peak of its trajectory, the highest point above the ground.

To find this maximum height:

• Use the vertical motion equation \( v_{y} = v_{0y} - gt \) to find when \( v_y = 0 \).
• Calculate the corresponding time (\( t \)) at this point using \( t = \frac{v_{0y}}{g} \).

Plug this time back into the vertical position equation:

• \[ y = y_0 + v_{0y} \cdot t - \frac{1}{2}gt^2 \]

This calculation gives you the exact highest point reached by the projectile. Knowing the maximum height is vital for applications where clearance above ground is essential, such as sports or aerospace disciplines.