Sci212e Physicsfor This Homework Carefully Study Examples 34 ✓ Solved

For this homework, carefully study Examples 3.4 and 3.5 from the textbook, found in the section 3.4, Projectile Motion: Now work on them with different initial velocity: v0 = 100 m/s at an angle 45o above horizon. Use g = -10.0 m/s as the gravitational acceleration instead of g = -9.8 m/s.

Work on the following problems:

[1] During a fireworks display, a shell is shot into the air with an initial speed of 100 m/s at an angle of 45.0º above the horizontal. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

[2] A large rock is ejected from Kilauea in Hawaii with a speed of 100 m/s and at an angle 45.0º above the horizontal. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and direction of the rock’s velocity at impact? Again, use g = -10.0 m/s. Please show all your work, not just the answers. Don’t forget the appropriate units for all cases.

Paper For Above Instructions

Projectile motion refers to the motion of an object that is thrown into the air and is subject to the force of gravity. This type of analysis is important in physical sciences, particularly physics and engineering. In this paper, we will discuss two scenarios involving projectile motion: the trajectory of a shell in a fireworks display and the ejection of a rock from the active volcano Kilauea.

Scenario 1: Fireworks Display

For the first scenario, we have a shell launched into the air with an initial velocity of v0 = 100 m/s at an angle of 45º above the horizontal. Applying the principles of projectile motion, we can calculate the height at which the shell explodes, the time taken to reach the maximum height, and the horizontal displacement at that moment.

1. Height of the Shell at Explosion

The maximum height (\(h\)) achieved by the projectile can be calculated using the formula:

\(h = \frac{v_{0}^{2} \sin^{2}(\theta)}{2 |g|}\)

Where:

  • \(v_{0} = 100 m/s\)

  • \(\theta = 45º\)

  • \(g = -10.0 m/s^{2}\) (taking the negative to indicate a downward direction)

Substituting the values:

\(h = \frac{(100)^{2} \sin^{2}(45º)}{2 \times 10}\)

\(h = \frac{10000 \times 0.5}{20} = 250 m\)

Thus, the shell explodes at a height of 250 meters.

2. Time to Reach Maximum Height

The time (\(t\)) taken to reach the maximum height can be calculated using the formula:

\(t = \frac{v_{0} \sin(\theta)}{|g|}\)

Substituting these values:

\(t = \frac{100 \sin(45º)}{10}\)

\(t = \frac{100 \times 0.7071}{10} \approx 7.07 s\)

Therefore, the time passed between the launch of the shell and the explosion is approximately 7.07 seconds.

3. Horizontal Displacement at Explosion

The horizontal displacement (\(d\)) when the shell explodes can be calculated with:

\(d = v_{0} \cos(\theta) \times t\)

Substituting these values:

\(d = 100 \cos(45º) \times 7.07\)

\(d = 100 \times 0.7071 \times 7.07 \approx 500 m\)

Thus, the horizontal displacement of the shell when it explodes is approximately 500 meters.

Scenario 2: Ejected Rock from Kilauea

In the second scenario, we have a rock ejected with the same initial speed of 100 m/s and at an angle of 45º above the horizontal. The rock strikes the side of the volcano at an altitude of 20 m lower than its starting point.

1. Time Taken for the Rock

To find the time taken for the rock to hit the side of the volcano, we can use the kinematic equations. The vertical displacement (\(y\)) is given as:

\(y = v_{0y} \times t + \frac{1}{2}(-g)t^{2}\)

Where \(v_{0y} = v_{0} \sin(\theta)\). Hence, we can rearrange the equation:

\(-20 = 100 \times 0.7071 \times t - 5t^{2}\)

Rearranging gives:

\(5t^{2} - 70.71t - 20 = 0\)

Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\), where \(a = 5, b = -70.71, c = -20\):

Solving gives \(t \approx 14.16seconds\) (taking positive root).

2. Velocity at Impact

To determine the velocity at impact, we calculate the vertical and horizontal components of the velocity just before it strikes the side of the volcano:

\(v_{x} = v_{0} \cos(\theta)\)

\(v_{y} = v_{0} \sin(\theta) - g \cdot t\)

Substituting the values:

\(v_{x} = 100 \cos(45º) = 70.71 m/s\)

\(v_{y} = 100 \sin(45º) - 10 \cdot 14.16\)

\(v_{y} = 70.71 - 141.6 = -70.89 m/s\)

The magnitude of the total velocity (\(v\)) is:

\(v = \sqrt{v_{x}^2 + v_{y}^2}\)

\(v = \sqrt{(70.71)^2 + (-70.89)^2} \approx 100.72 m/s\)

The direction can be calculated using \(\theta_{impact} = \tan^{-1}\left(\frac{v_{y}}{v_{x}}\right) \approx -45.1º\). Thus, at impact, the rock has a velocity of approximately 100.72 m/s at an angle of 45.1º below the horizontal.

Conclusion

Through this analysis of the fireworks shell and the ejected rock from Kilauea, we successfully calculated various aspects of projectile motion using varying initial velocities and gravitational forces.

References

  • Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics. Wiley.

  • Serway, R. A., & Jewett, J. W. (2014). Physics for Scientists and Engineers. Cengage Learning.

  • Young, H. D., & Freedman, R. A. (2014). University Physics with Modern Physics. Pearson.

  • Tipler, P. A., & Mosca, G. (2007). Physics for Scientists and Engineers. W.H. Freeman.

  • Giancoli, D. C. (2013). Physics: Principles with Applications. Pearson.

  • Harris, E. W. (2011). Physics: A Brief History. Cambridge University Press.

  • Ryder, F. (2007). Introduction to Classical Mechanics. McGraw-Hill.

  • Wolfson, R. (2012). Essentials of Physics. Addison-Wesley.

  • Resnick, R., & Halliday, D. (2000). Physics (5th ed.). Wiley.

  • Purcell, E. M., & Morin, D. (2013). Electricity and Magnetism. Cambridge University Press.