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Chapter 13: Problem 3

Flight time and height A projectile is fired with an initial speed of 500 \(\mathrm{m} / \mathrm{sec}\) at an angle of elevation of \(45^{\circ} .\) a. When and how far away will the projectile strike? b. How high overhead will the projectile be when it is 5 \(\mathrm{km}\) downrange? c. What is the greatest height reached by the projectile?

Short Answer

Expert verified
a. The projectile lands at 25,500 m after 72.15 sec. b. At 5 km, the height is 4244 m. c. Maximum height is 6364 m.

Step by step solution

01

Resolve Initial Velocity Components

First, we need to find the components of the initial velocity. The projectile has an initial velocity of 500 m/s at a 45掳 angle to the horizontal. Use trigonometry: \( v_{x_0} = v_0 \cos(\theta) \) and \( v_{y_0} = v_0 \sin(\theta) \), where \( v_0 = 500\, \mathrm{m/s} \) and \( \theta = 45^{\circ} \). Thus, \( v_{x_0} = 500 \cos(45^{\circ}) = 353.55\, \mathrm{m/s} \) and \( v_{y_0} = 500 \sin(45^{\circ}) = 353.55\, \mathrm{m/s} \).
02

Determine Time of Flight

The time of flight can be found using the vertical motion equation, knowing that the vertical velocity at peak and landing is zero: \( v_y = v_{y_0} - gt \). Because \( v_y = -v_{y_0} \) when the projectile lands, set \( v_{y_0} - g \cdot t = -v_{y_0} \). Solving gives \( t = \frac{2 v_{y_0}}{g} = \frac{2 \times 353.55}{9.8} \approx 72.15\, \mathrm{sec} \).
03

Calculate Range

The range (horizontal distance) can be found using \( x = v_{x_0} \times t \). Plug in \( v_{x_0} = 353.55\, \mathrm{m/s} \) and \( t = 72.15\, \mathrm{sec} \): \( x \approx 353.55 \times 72.15 \approx 25500\, \mathrm{meters} \).
04

Find the Height at 5 km Downrange

At \( 5 \mathrm{km} \) (or 5000 meters) downrange, calculate the time \( t \) to reach 5 km: \( t = \frac{5000}{353.55} \approx 14.14\, \mathrm{sec} \). The height during this time is \( y = v_{y_0}t - \frac{1}{2}gt^2 = 353.55 \times 14.14 - 0.5 \times 9.8 \times (14.14)^2 \approx 4244 \mathrm{meters} \).
05

Calculate Maximum Height

The maximum height is reached when the vertical component of the velocity is zero. Use the equation \( v_y = v_{y_0} - gt \), setting \( v_y = 0 \): \( t = \frac{v_{y_0}}{g} = \frac{353.55}{9.8} \approx 36.075\, \mathrm{sec} \). Now, calculate the height: \( h_{max} = v_{y_0}t - \frac{1}{2}gt^2 = 353.55 \times 36.075 - 0.5 \times 9.8 \times (36.075)^2 \approx 6364\, \mathrm{meters} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Velocity Components
When dealing with projectile motion, the first step is to understand the initial velocity components. These are crucial because they set the foundation for figuring out how far and how high a projectile will fly. For a projectile launched at an angle, the velocity is split into two components: one horizontal
  • The horizontal component, \( v_{x_0} = v_0 \cos(\theta) \), determines how fast the projectile moves along the horizontal plane.
  • The vertical component, \( v_{y_0} = v_0 \sin(\theta) \), impacts how high the projectile will rise.
In the provided problem, an initial speed of 500 m/s at a 45掳 angle gives us equal components due to the angle's properties, resulting in both 353.55 m/s horizontally and vertically.
Understanding these separate vectors helps break down projectile motion into two simpler, linear motions鈥攈orizontal and vertical.
Time of Flight
The time of flight indicates how long a projectile remains in the air. Calculating it involves considering the vertical motion since gravity exclusively influences this movement. The key step is to find when the vertical velocity hits zero and then doubles to account for the ascent and descent. In mathematical terms:
  • Start with the equation of motion: \( v_y = v_{y_0} - gt \)
  • For total flight, set final vertical velocity (\( v_y \)) to \(-v_{y_0} \)
  • This rearranges to \( t = \frac{2v_{y_0}}{g} \)
  • Plug in the values for our example, \( v_{y_0} = 353.55\, \text{m/s}\), \( g = 9.8\, \text{m/s}^2\), yielding a time of flight of approximately 72.15 seconds.
This total time allows us to further calculate distances and heights in the projectile's trajectory.
Range Calculation
The range of the projectile refers to the total horizontal distance it travels before hitting the ground. Calculated using the horizontal component of initial velocity and the time of flight, the formula used is simple yet highly insightful:
  • Use \( x = v_{x_0} \times t \), where \( x \) is the range.
  • For the example, \( v_{x_0} = 353.55\, \text{m/s}\) and the flight time \( t = 72.15\, \text{sec} \)
  • Calculating this: \( x = 353.55 \times 72.15 \approx 25500\, \text{meters} \)
Understanding and applying these calculations not only solve for the range but deepen your grasp of how different variables affect projectile motion.
Maximum Height
The maximum height is perceived at the peak of the projectile's arc where the vertical velocity momentarily ceases. Calculating this involves analyzing when the upward velocity dips to zero. Here's how you solve it:
  • Use \( v_y = v_{y_0} - gt \) and set \( v_y \) to zero for the peak point
  • This translates into \( t_{peak} = \frac{v_{y_0}}{g} \)
  • Substitute to find the peak time: \( t_{peak} = \frac{353.55}{9.8} \approx 36.075\, \text{sec} \)
  • Use \( h_{max} = v_{y_0}t_{peak} - \frac{1}{2}gt_{peak}^2 \) to find the peak height, calculating to about \( 6364\, \text{meters} \)
This calculation emphasizes the symphony of physics in motion and demonstrates how speed, angle, and gravity interplay to govern the flight path.
Physics Problem Solving
Projectile motion problems can appear daunting, but breaking them down into calculated steps can simplify the process. Remember to:
  • Start with identifying the known quantities and assign variables.
  • Use trigonometric relationships for resolving initial velocities.
  • Apply kinematic equations for time of flight, range, and maximum height sequentially.
  • Verify each step with the context of the problem - does it make sense practically?
Each physics problem is a puzzle; understand the principles, apply the formulas accurately, and you will develop an intuitive feel for how objects move in various forces. Once you grasp these basics, you'll feel more comfortable approaching any projectile motion scenario confidently.

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