Problem 55 - Calculus Volume 3

Question

The trajectory of a bullet is given by \(x=v_{0}(\cos \alpha) t y=v_{0}(\sin \alpha) t-\frac{1}{2} g t^{2}\) where \(v_{0}=500 \mathrm{~m} / \mathrm{s}, \quad g=9.8=9.8 \mathrm{~m} / \mathrm{s}^{2}, \quad\) and \(\alpha=30\) degrees. When will the bullet hit the ground? How far from the gun will the bullet hit the ground?

Step-by-Step Solution

Answer

The bullet hits the ground after approximately 51.02 seconds, 22,107 meters away.

Step 1: Understanding the Problem

We need to determine when the bullet hits the ground and the distance from the gun at that point. This involves solving for the time at which the bullet hits the ground and using that time to find the horizontal distance.

Step 2: Calculate Time of Flight

The bullet hits the ground when its height, given by the equation \(y=v_{0}( an \alpha) t - \frac{1}{2}g t^{2}\), is equal to zero. We solve for time by setting \(y = 0\): \[0 = 500(\sin 30^{\circ})t - \frac{1}{2} \times 9.8 \times t^{2}\]Simplifying, we have:\[0 = 250t - 4.9t^{2}\]\[t(250 - 4.9t) = 0\]This gives us \(t = 0\) or \(t = \frac{250}{4.9}\). Since \(t = 0\) is the initial time, the bullet hits the ground at \(t = \frac{250}{4.9}\). Solving, \(t \approx 51.02\, \text{seconds}\).

Step 3: Calculate Horizontal Distance

Using the time of flight (\(t \approx 51.02\, \text{seconds}\)), calculate the horizontal distance \(x\). The equation for \(x\) is given by:\[x = v_{0}(\cos \alpha) t = 500(\cos 30^{\circ})(51.02)\approx 500(0.866)(51.02)\]Thus, \(x \approx 22106.87\) meters.

Key Concepts

Time of Flight

In projectile motion, the time of flight is the duration from when the projectile is launched until it hits the ground again. For a projectile launched at an angle \( \alpha \) with initial velocity \( v_0 \), we can determine the time of flight using the equation of motion in the vertical direction. The projectile returns to the ground when its vertical displacement \( y \) is zero.

• The formula for vertical motion is \( y = v_{0} ( \sin \alpha ) t - \frac{1}{2} g t^2 \).
• Setting \( y = 0 \), we solve the equation to find \( t = \frac{2v_{0}(\sin \alpha)}{g} \).
• For this problem, with \( \alpha = 30^{\circ} \), \( v_{0} = 500 \) m/s, and \( g = 9.8 \text{ m/s}^2 \), solving the equation gives \( t \approx 51.02 \) seconds.

Recognizing that time of flight depends on both the launch angle and the initial velocity helps understand why the bullet travels for such an extended period.

Horizontal Distance

Horizontal distance, often called "range", is how far the projectile travels horizontally before hitting the ground. The distance depends on the initial speed, angle of launch, and time of flight.

• The horizontal distance formula is \( x = v_{0}(\cos \alpha) t \).
• This equation assumes zero air resistance, allowing for straightforward calculation using time from the time of flight formula.
• Substituting the time we found, \( t \approx 51.02 \) seconds, into the horizontal distance formula with \( v_{0} = 500 \) m/s and \( \alpha = 30^{\circ} \) results in \( x \approx 22,106.87 \) meters.

Understanding this concept enables us to predict how far a projectile will go based on its initial conditions, allowing for practical applications in various fields like ballistics and sports.

Trajectory Equation

The trajectory equation is essential for describing the path of a projectile. It provides a mathematical representation of the projectile's motion through a set of parametric equations for horizontal and vertical positions as functions of time.

• Horizontal motion: \( x = v_{0}(\cos \alpha) t \).
• Vertical motion: \( y = v_{0}(\sin \alpha) t - \frac{1}{2} g t^2 \).

These parametric equations detail how the projectile's velocity components influence its path and ultimate trajectory.
While solving such equations, it is crucial to recognize how each component individually and collectively contributes to the projectile's motion. By examining these factors, one can effectively handle a wide variety of physics problems involving projectile motion.

Physics Problem Solving

Physics problem solving requires a methodical approach to understand and solve complex problems. Specifically, for projectile motion problems like the one provided, this process generally involves several steps:

• First, interpret the problem: Identify what is given and what needs to be solved.
• Second, apply relevant physics equations: Use equations for motion such as the trajectory equations for both x and y to relate variables logically.
• Third, solve the mathematics: Rearrange and solve equations for unknowns, making sure to use correct units and consistent values.
• Lastly, interpret the result: Ensure the solution makes sense in the physical context and checks for any approximations or assumptions.

By following these steps, one can be confident in tackling various physics problems, enhancing both understanding and problem-solving skills for real-world applications.