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 \(\mathbf{v}_{0}=\left\langle u_{0}, v_{0}\right\rangle\). $$\left\langle u_{0}, v_{0}\right\rangle=(40,80) \mathrm{m} / \mathrm{s}$$

Short answer

Question: For a two-dimensional projectile trajectory with an initial velocity vector \(\mathbf{v}_{0} = (40,80) \mathrm{m}/\mathrm{s}\), find the time of flight, range, and maximum height. Answer: For the given trajectory, the time of flight is approximately \(11.79 \mathrm{s}\), the range is approximately \(318.83 \mathrm{m}\), and the maximum height is approximately \(204.08 \mathrm{m}\).

Step-by-step solution

Identify the known values and relevant formulas

The known values are: 1. Initial position: \((0,0)\) 2. Initial velocity vector: \(\mathbf{v}_{0} = \left\langle u_{0}, v_{0}\right\rangle = (40,80) \mathrm{m}/\mathrm{s}\) 3. Acceleration due to gravity: \(g = 9.81 \mathrm{m}/\mathrm{s}^{2}\) (downwards) The following equations will be used to solve the problem: 1. Time of flight: \(t_{flight} = \dfrac{2v_0 \sin{\theta}}{g}\) 2. Range: \(R = \dfrac{v_0^2 \sin{2\theta}}{g}\) 3. Maximum height: \(H_{max} = \dfrac{v_0^2 \sin^2{\theta}}{2g}\)

Calculate the angle \(\theta\)

The angle \(\theta\) in the initial velocity vector can be calculated using the tangent function: \(\tan{\theta} = \dfrac{v_0}{u_0} = \dfrac{80}{40} = 2\) Now we find the angle \(\theta\): \(\theta = \arctan(2) \approx 63.4^{\circ}\)

Calculate the time of flight \(t_{flight}\)

Using the time of flight formula: \(t_{flight} = \dfrac{2v_0 \sin{\theta}}{g} = \dfrac{2 \times 90 \times \sin{63.4^{\circ}}}{9.81} \approx 11.79 \mathrm{s}\)

Calculate the range \(R\)

Using the range formula: \(R = \dfrac{v_0^2 \sin{2\theta}}{g} = \dfrac{90^2 \times \sin{2 \times 63.4^{\circ}}}{9.81} \approx 318.83 \mathrm{m}\)

Calculate the maximum height \(H_{max}\)

Using the maximum height formula: \(H_{max} = \dfrac{v_0^2 \sin^2{\theta}}{2g} = \dfrac{90^2 \times \sin^2{63.4^{\circ}}}{2 \times 9.81} \approx 204.08 \mathrm{m}\) In conclusion, for the given two-dimensional trajectory with an initial velocity vector \(\mathbf{v}_{0} = (40,80) \mathrm{m}/\mathrm{s}\), the time of flight is approximately \(11.79 \mathrm{s}\), the range is approximately \(318.83 \mathrm{m}\), and the maximum height is approximately \(204.08 \mathrm{m}\).

Key concepts

Time of Flight

The time of flight of a projectile refers to the total time a projectile is in motion from the moment it is launched until the moment it returns to the ground. Understanding the time of flight is crucial for predicting where a projectile will land. To find the time of flight, one can use the formula:\[t_{\text{flight}} = \dfrac{2v_0 \sin{\theta}}{g}\]where:

• \(v_0\) is the initial velocity magnitude,
• \(g\) is the acceleration due to gravity \((9.81 \text{m/s}^2)\),
• \(\theta\) is the launch angle relative to the horizontal axis.

In this scenario, we first need to determine the angle \(\theta\) using the tangent function: \[\tan{\theta} = \dfrac{v_0}{u_0} = \dfrac{80}{40} = 2\]which gives \(\theta \approx 63.4°\). Now, substituting \(\theta\) back into the time of flight formula, we find that the time of flight is approximately \(11.79 \text{ seconds}\). This means the projectile will be airborne for this duration until it hits the ground again.

Range of Projectile

The range of a projectile is the horizontal distance it travels while it is in motion. For many practical applications, determining how far a projectile will go is useful. The formula for calculating the range of a projectile is:\[R = \dfrac{v_0^2 \sin{2\theta}}{g}\]where:

• \(v_0\) stands for the initial velocity magnitude,
• \(\theta\) is the launch angle,
• \(g\) is the acceleration due to gravity \((9.81 \text{m/s}^2)\).

Remember, \(\sin{2\theta}\) is a critical component of this calculation, derived from the angle \(\theta\) we computed earlier. By plugging our known values, including \(\theta ≈ 63.4°\), into the range formula, we get the range \(R \approx 318.83 \text{ meters}\). This figure reflects the horizontal distance covered by the projectile during its flight.

Maximum Height

The maximum height of a projectile indicates the highest point it reaches along its trajectory. Understanding this concept is essential when ensuring a projectile clears obstacles or reaches a certain height. The formula to calculate the maximum height is:\[H_{\text{max}} = \dfrac{v_0^2 \sin^2{\theta}}{2g}\]where:

• \(v_0\) is the initial velocity magnitude,
• \(g\) is gravity, at approximately \(9.81 \text{m/s}^2\),
• \(\theta\) is the launch angle.

In our scenario, knowing \(\theta \approx 63.4°\) and substituting into the maximum height formula, we find that the projectile reaches a maximum height of about \(204.08 \text{ meters}\).This calculation is particularly useful for determining if a projectile can clear specific vertical barriers during its flight.