Question

Find the time of flight, range, and maximum height of the following two- dimensional trajectories, assuming no forces other than gravity. In each case the initial position is (0,0) and the initial velocity is \(\mathbf{v}_{0}=\left\langle u_{0}, v_{0}\right\rangle\). $$\text { Initial speed }\left|\mathbf{v}_{0}\right|=400 \mathrm{ft} / \mathrm{s}, \text { launch angle } \alpha=60^{\circ}$$

Short answer

Answer: The time of flight is 21.57 seconds, the range is 4314 ft, and the maximum height is approximately 1871 ft.

Step-by-step solution

Determine Initial Velocity Components

To break down the initial velocity vector into x and y components, we use the following equations: $$v_{0x} = |\mathbf{v}_{0}| \cos \alpha$$ $$v_{0y} = |\mathbf{v}_{0}| \sin \alpha$$ Given \(|\mathbf{v}_{0}| = 400\ \text{ft/s}\) and \(\alpha = 60^{\circ}\), we can find \(v_{0x}\) and \(v_{0y}\) as: $$v_{0x} = 400\ \text{ft/s} \cos 60^{\circ} = 200\ \text{ft/s}$$ $$v_{0y} = 400\ \text{ft/s} \sin 60^{\circ} = 400\ \text{ft/s} * \frac{\sqrt{3}}{2} = 346.41\ \text{ft/s}$$

Calculate Time of Flight

The time of flight is the time it takes for the projectile to reach the ground. Since vertical motion is affected by gravity, we use the equation for the y-component of motion to find the time of flight: $$y(t) = v_{0y}t - \frac{1}{2}gt^2$$ At the landing point, the final vertical position is y = 0. So, we can find the time of flight by setting y(t) = 0 and solving for t: $$0 = 346.41\ \text{ft/s} * t - \frac{1}{2}(32.174\ \text{ft/s}^2) * t^2$$ There are two possible solutions for t: one when the projectile is at its starting point, and another when it lands back on the ground. The first solution is t = 0, which corresponds to the starting time, and the second solution is the time of flight. By solving the quadratic equation for t, we find the time of flight to be \(t_f = 21.57\ \text{s}\).

Calculate Range

The range of the projectile is the horizontal distance it travels during its time of flight. Using the horizontal component of the projectile's motion, we can find the range as follows: $$x(t) = v_{0x}t$$ Substituting the time of flight (\(t_f = 21.57\ \text{s}\)) and the horizontal component of the initial velocity (\(v_{0x} = 200\ \text{ft/s}\)), we get: $$x(t_f) = 200\ \text{ft/s} * (21.57\ \text{s}) = 4314\ \text{ft}$$ So the range is 4314 ft.

Calculate Maximum Height

The maximum height is achieved when the vertical component of the projectile's velocity becomes zero (i.e., \(v_y = 0\)). We use the following equation to find the time it takes to reach the maximum height: $$v_y = v_{0y} - gt$$ Setting \(v_y = 0\) and solving for t: $$0 = 346.41\ \text{ft/s} - (32.174\ \text{ft/s}^2) * t$$ $$t_{max} = \frac{346.41\ \text{ft/s}}{32.174\ \text{ft/s}^2} = 10.785\ \text{s}$$ Now that we have found the time at which the maximum height occurs, we can use the y-component equation to find the maximum height itself: $$y(t_{max}) = 346.41\ \text{ft/s} * 10.785\ \text{s} - \frac{1}{2}(32.174\ \text{ft/s}^2)(10.785\ \text{s})^2$$ $$y(t_{max}) \approx 1871\ \text{ft}$$ So the maximum height is approximately 1871 ft. In conclusion, for this two-dimensional projectile motion with initial velocity \(\mathbf{v}_{0}=\left\langle u_{0}, v_{0}\right\rangle\) and initial speed \(|\mathbf{v}_{0}|=400\ \text{ft/s}\) at a launch angle \(\alpha=60^{\circ}\), the time of flight is \(21.57\ \text{s}\), the range is \(4314\ \text{ft}\), and the maximum height is approximately \(1871\ \text{ft}\).

Key concepts

Time of Flight

The "time of flight" in projectile motion refers to the total time that the projectile is in the air. During projectile motion, the only force acting on the object is gravity, which affects the vertical components of the motion. To determine the time of flight, we utilize the kinematic equation for vertical motion:

• \( y(t) = v_{0y} \cdot t - \frac{1}{2}gt^2 \)

The equation sets the vertical position \( y(t) \) to zero at landing. By solving this for \( t \), we can calculate the time of flight. This step involves solving a quadratic equation as the projectile travels from its initial vertical displacement up to its peak and back to the horizontal plane. Using the values given: \( v_{0y} = 346.41\ \text{ft/s} \) and gravitational acceleration \( g = 32.174\ \text{ft/s}^2 \), the time of flight becomes 21.57 seconds. This calculation ensures that each segment of the flight—from takeoff to landing—is accounted for in the total time.

Range

The "range" of a projectile is the horizontal distance it covers during its time of flight. To calculate this, we focus on the horizontal component of motion. This component remains unaffected by gravity and moves with a constant velocity. We can determine range using:

• \( x(t) = v_{0x} \cdot t_{f} \)

Here, \( v_{0x} \) is the horizontal component of the initial velocity, and \( t_{f} \) is the total time of flight. Given \( v_{0x} = 200\ \text{ft/s} \), with a flight time of 21.57 seconds, the range calculates to 4314 feet. This tells us how far along the horizontal axis the projectile traveled while air-bound, aligning with real-world applications such as determining how far an object can be thrown or launched.

Maximum Height

The "maximum height" of a projectile is the highest point reached in its trajectory. At this peak, the vertical velocity equals zero momentarily before the projectile descends. Computing the maximum height requires us to calculate the time it takes to reach this point, then use it to find the vertical height. To find the time to maximum height \( t_{max} \):

• \( v_y = v_{0y} - gt \rightarrow 0 = 346.41 - 32.174 \cdot t \)

Solving for \( t \), we find \( t_{max} = 10.785\ \text{s} \). We then use the equation for vertical displacement to find maximum height:

• \( y(t_{max}) = v_{0y} \cdot t_{max} - \frac{1}{2}gt_{max}^{2} \)

This calculates to approximately 1871 feet, indicating the apex for this projectile motion.

Initial Velocity Components

Understanding the "initial velocity components" is crucial in dissecting a projectile's motion. By dividing the initial velocity into horizontal and vertical components, we simplify the analysis and calculations. The initial velocity vector is expressed as:

• \( \mathbf{v}_{0} = \langle u_{0}, v_{0} \rangle \)

Using trigonometry, we separate this vector based on the launch angle \( \alpha \).

• Horizontal component: \( v_{0x} = |\mathbf{v}_{0}| \cos \alpha \)
• Vertical component: \( v_{0y} = |\mathbf{v}_{0}| \sin \alpha \)

For an initial speed of 400 ft/s and a launch angle of 60°, these components compute to \( v_{0x} = 200\ \text{ft/s} \) and \( v_{0y} = 346.41\ \text{ft/s} \). These components help determine each aspect of the projectile's journey, ensuring accurate calculations for time of flight, range, and maximum height.