Question

Trajectory properties 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 \(\mathbf{v}_{0}=\left\langle u_{0}, v_{0}\right\rangle\). Initial speed \(\left|\mathbf{v}_{0}\right|=150 \mathrm{m} / \mathrm{s},\) launch angle \(\alpha=30^{\circ}\)

Short answer

Question: A projectile is launched with an initial speed of 150 m/s at an angle of 30 degrees. Calculate the time of flight, horizontal range, and maximum height. Answer: For the given projectile motion, the time of flight is approximately 15.29 seconds, the horizontal range is approximately 1980.11 meters, and the maximum height is approximately 284.84 meters.

Step-by-step solution

Calculate the initial horizontal and vertical components of velocity

To solve this problem, we need to first find the horizontal and vertical components of the initial velocity. Since the initial speed is 150 m/s and the launch angle is 30 degrees, we can use the following formulas: \(u_{0} = |\mathbf{v}_{0}| \cos{\alpha}\) - horizontal component of velocity. \(v_{0} = |\mathbf{v}_{0}| \sin{\alpha}\) - vertical component of velocity. Using the given values, we get: \(u_{0} = 150 \cos{30^{\circ}} = 150 \times \frac{\sqrt{3}}{2} = 75\sqrt{3} \text{ m/s}\) \(v_{0} = 150 \sin{30^{\circ}} = 150 \times \frac{1}{2} = 75 \text{ m/s}\)

Calculate the time of flight

The time of flight (\(t_{f}\)) can be calculated using the vertical component of the initial velocity, \(v_{0}\), and the formula: \(t_{f} = \frac{2v_{0}}{g}\), where \(g\) is the acceleration due to gravity (approximately 9.81 m/s²). Substituting the given values, we get: \(t_{f} = \frac{2 \times 75}{9.81} \approx 15.29 \text{ s}\) The time of flight is approximately 15.29 seconds.

Calculate the horizontal range

The horizontal range (\(R\)) can be calculated using the time of flight, \(t_{f}\), and the horizontal component of the initial velocity, \(u_{0}\), with the formula: \(R = u_{0} t_{f}\) Substituting the given values, we get: \(R = 75\sqrt{3} \times 15.29 \approx 1980.11 \text{ m}\) The horizontal range is approximately 1980.11 meters.

Calculate the maximum height

The maximum height (\(H_{max}\)) can be calculated using the vertical component of the initial velocity, \(v_{0}\), and the formula: \(H_{max} = \frac{v_{0}^{2}}{2g}\), where \(g\) is the acceleration due to gravity. Substituting the given values, we get: \(H_{max} = \frac{75^2}{2 \times 9.81} \approx 284.84 \text{ m}\) The maximum height is approximately 284.84 meters. So, for the given projectile motion with an initial speed of 150 m/s at 30 degrees launch angle, the time of flight is approximately 15.29 seconds, the horizontal range is approximately 1980.11 meters, and the maximum height is approximately 284.84 meters.

Key concepts

Time of Flight

The time of flight in projectile motion refers to the total duration a projectile is in the air from the moment it is launched until it hits the ground. This measure is critically important because it directly impacts other characteristics of the projectile's trajectory, such as its horizontal range and maximum height.

For a projectile launched at an angle with the horizontal, the time of flight depends on the vertical component of the initial velocity and the acceleration due to gravity. The formula given by
\[ t_f = \frac{2v_0}{g} \]
assumes the starting and ending heights are the same, which is the case in many introductory physics problems. Using the formula with our example's vertical velocity component of 75 m/s and gravity at approximately 9.81 m/s², we've determined a time of flight of around 15.29 seconds. It is essential to note that this duration would change if the projectile was launched from a height or aimed toward a different ending elevation.

Horizontal Range

The horizontal range of a projectile is the total horizontal distance traveled from its point of launch until it hits the ground. The way to find this distance is fairly intuitive; it is the product of the horizontal component of the initial velocity and the time the projectile spends in the air (time of flight).

The relevant formula for our example is
\[ R = u_0 t_f \]
where we insert the horizontal component of velocity (calculated using trigonometry) and the previously computed time of flight. By multiplying 75\(\sqrt{3}\) m/s by 15.29 seconds, we calculated the projectile's range to be about 1980.11 meters. Understanding the horizontal range is crucial for applications such as sports, artillery, and predicting where a space vehicle will land.

Maximum Height

Achieving the maximum height is when the projectile's vertical velocity becomes zero before it starts descending again due to gravity. The maximum height is significant because it tells us how high a projectile will rise, which is critical in scenarios where clearances or obstacles must be considered.

To find the maximum height, use the formula
\[ H_{max} = \frac{v_0^2}{2g} \]
where we take the initial vertical velocity and square it, then divide by twice the acceleration due to gravity. We determined the maximum height to be around 284.84 meters using the starting vertical velocity of 75 m/s from our example. This calculation has immense importance in fields such as aeronautics and engineering where altitude is a deciding factor for successful projectile deployment.

Initial Velocity Components

Understanding a projectile's initial velocity components is essential because they shape its trajectory. These components are found by breaking down the initial velocity into horizontal (\(u_0\)) and vertical (\(v_0\)) parts.

Using trigonometric functions, we obtain
\[ u_0 = |\mathbf{v}_0| \cos(\alpha) \]
\[ v_0 = |\mathbf{v}_0| \sin(\alpha) \]
In our example, the initial velocity was broken down into a horizontal component of 75\(\sqrt{3}\) m/s and a vertical component of 75 m/s. These components are crucial to the overall behavior of the projectile; for example, the horizontal component affects the range, and the vertical component affects the time of flight and maximum height. Therefore, accurately identifying these components is the foundation for all other calculations in projectile motion.

Acceleration Due to Gravity

The acceleration due to gravity, often symbolized by\(g\), is constant at Earth's surface and is approximately 9.81 m/s². This acceleration governs the motion of all freely-falling objects and, in projectile motion, it only affects the vertical motion of the projectile.

Gravity is what pulls a projectile back to Earth's surface and determines how long it stays in the air and how high it can reach. Every calculation in projectile motion pertaining to time of flight or maximum height requires an accurate value of\(g\), and all force except gravity is usually neglected in basic physics problems. Remember that gravity's influence is global, but its exact value can differ slightly depending on one's location on Earth—something that could be vital for precision tasks in engineering or space exploration.