For This Homework: Carefully Study Examples 34 And 35 396936

For This Homework Carefully Study Examples 34 And 35 From The Textb

Carefully study Examples 3.4 and 3.5 from the textbook, found in section 3.4, Projectile Motion. Then, work on these problems with a different initial velocity: v₀ = 100 m/s at an angle of 45° above the horizontal. Use g = -10.0 m/s² for gravitational acceleration instead of -9.8 m/s². The specific problems are:

1. During a fireworks display, a shell is shot into the air with an initial speed of 100 m/s at an angle of 45.0° above the horizontal. The fuse ignites the shell just as it reaches its highest point. Calculate:
   • (a) the height at which the shell explodes;
   • (b) the time elapsed from launch to explosion;
   • (c) the horizontal displacement of the shell when it explodes.
2. Kilauea volcano in Hawaii ejects rocks at a speed of 100 m/s at an angle of 45.0° above the horizontal. A large rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. Calculate:
   • (a) the time it takes for the rock to follow this path;
   • (b) the magnitude and direction of the rock’s velocity at impact.

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Paper For Above instruction

Introduction

Projectile motion is a fundamental concept in physics, describing the trajectory of objects subjected to gravity after being given an initial velocity. Understanding the kinematic equations that govern these motions allows us to analyze and predict the behaviors of projectiles in various scenarios. The problems discussed herein seek to apply the principles of projectile motion with an initial velocity of 100 m/s at a 45° launch angle, utilizing a gravitational acceleration of -10.0 m/s², a slight deviation from the standard -9.8 m/s². These cases include a fireworks shell reaching its highest point, and a volcanic rock descending to a specified lower elevation on a volcano's side, illustrating key concepts such as maximum height, time of flight, and impact velocity.

Problem 1: Fireworks Shell at Its Highest Point

Background and Known Data

The initial velocity v₀ = 100 m/s is launched at an angle θ = 45°. The acceleration due to gravity g = -10.0 m/s² acts downward. When the fuse ignites the shell at maximum height, the vertical component of velocity (v_y) is zero, signifying the peak of its trajectory.

Calculations

(a) Height at Explosion

The vertical component of initial velocity is:

v_{0y} = v₀ sin θ = 100 sin 45° ≈ 100 * 0.7071 ≈ 70.71 m/s

Using the kinematic equation for vertical displacement:

v_y² = v_{0y}² + 2g * h

At maximum height, v_y = 0, so:

0 = (70.71)² + 2 (-10) h

h = (70.71)² / (2 * 10) = 5000 / 20 = 250 m

(b) Time to Reach Maximum Height

Time taken to reach maximum height (as vertical velocity becomes zero):

t_{up} = v_{0y} / |g| = 70.71 / 10 ≈ 7.07 s

Since this is the ascent phase and assuming symmetry, the total time of flight is twice this value:

t_{total} = 2 * 7.07 ≈ 14.14 s

The explosion occurs at the maximum height, so time elapsed from launch to explosion is approximately 7.07 seconds.

(c) Horizontal Displacement at Explosion

The horizontal component of initial velocity:

v_{0x} = v₀ cos θ = 100 0.7071 ≈ 70.71 m/s

Horizontal displacement (range at maximum height):

x = v_{0x} t_{up} ≈ 70.71 7.07 ≈ 500 m

Problem 2: Falling Rock on Volcano

Background and Known Data

The rock is ejected with v₀ = 100 m/s at 45°, landing 20 m below the starting point (initial height set as zero). The target is an altitude difference of -20 m. Using g = -10.0 m/s², the goal is to find the time of flight and impact velocity.

Calculations

(a) Time for the Path

Vertical component of initial velocity:

v_{0y} = 100 * sin 45° ≈ 70.71 m/s

Vertical displacement from start to impact (h): -20 m.

Using the vertical motion equation:

h = v_{0y} t + 0.5 g * t²

-20 = 70.71 t + 0.5 (-10) * t²

-20 = 70.71 t - 5 t²

Rearranged as:

5 t² - 70.71 t - 20 = 0

Dividing through by 5:

t² - 14.142 t - 4 = 0

Applying quadratic formula:

t = [14.142 ± √(14.142² - 4 1 (-4))] / 2

Discriminant D = 200 + 16 = 216

√D ≈ 14.7

Two solutions:

t = [14.142 + 14.7] / 2 ≈ 28.84 / 2 ≈ 14.42 s

or

t = [14.142 - 14.7] / 2 ≈ -0.558 / 2 ≈ -0.28 s (discard negative time)

Therefore, the time to impact is approximately 14.42 seconds.

(b) Velocity at Impact

Vertical component at impact:

v_y = v_{0y} + g t = 70.71 + (-10) 14.42 ≈ 70.71 - 144.2 ≈ -73.49 m/s

Horizontal component remains constant:

v_x = v_{0x} = 70.71 m/s

Magnitude of impact velocity:

v_{impact} = √(v_x² + v_y²) ≈ √(70.71² + (-73.49)²) ≈ √(5000 + 5401) ≈ √10,401 ≈ 102.0 m/s

Direction (angle below horizontal):

θ = arctangent (|v_y| / v_x) = arctangent (73.49 / 70.71) ≈ arctangent (1.04) ≈ 46.2°

The velocity vector at impact is approximately 102.0 m/s directed at 46.2° below the horizontal.

Conclusion

By applying corrected kinematic equations with the specified initial velocity and gravitational acceleration, we accurately determine the maximum height, time to reach that height, horizontal displacement at explosion, and impact parameters for the falling rock scenario. These calculations reinforce the critical understanding of projectile motion, showcasing the dependency on initial velocity components and acceleration due to gravity. Such analysis aids in predicting real-world projectile behaviors, as exemplified by fireworks displays and volcanic ejecta.

References

• Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers (9th ed.). Cengage Learning.
• Halliday, D., Resnick, R., & Walker, J. (2014). Fundamentals of Physics (10th ed.). Wiley.
• Young, H. D., & Freedman, R. A. (2019). University Physics (14th ed.). Pearson.
• Giancoli, D. C. (2013). Physics: Principles with Applications (7th ed.). Pearson.
• Tipler, P. A., & Mosca, G. (2008). Physics for Scientists and Engineers (6th ed.). W. H. Freeman.