For This Homework Carefully Study Examples 34 And 35

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

For this homework, carefully study Examples 3.4 and 3.5 from the textbook, found in section 3.4, Projectile Motion. Now work on them with different initial velocity: v0 = 100 m/s at an angle of 45º above the horizontal. Use g = -10.0 m/s² as the gravitational acceleration instead of g = -9.8 m/s². The problems are as follows:

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) Determine the time passed between the launch of the shell and the explosion. (c) Find the horizontal displacement of the shell when it explodes.
2. Kilauea in Hawaii is the world’s most continuously active volcano. Very active volcanoes eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 100 m/s at an angle of 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) Determine the magnitude and direction of the rock’s velocity at impact.

Show all your calculations with proper units for each step. You may submit your work using .doc, .docx, or .pdf files, or by taking pictures of handwritten work.

Paper For Above instruction

The study of projectile motion provides crucial insights into the trajectories of objects influenced solely by gravity, a fundamental aspect of physics with applications ranging from fireworks to natural phenomena like volcanic eruptions. This paper explores the calculations and analysis related to two distinct problems involving projectile motion, adapted to specific initial conditions: an exploding firework shell and a volcanic rock trajectory, both characterized by initial velocities of 100 m/s at a 45° launch angle, but with differing gravitational assumptions and contextual parameters.

Introduction

Projectile motion involves analyzing objects launched into the air under the influence of gravity, neglecting air resistance. The basic equations governing this motion depend on initial velocity, launch angle, acceleration due to gravity, and the time elapsed. The problems selected for this discussion involve calculating the maximum height, time of flight, horizontal displacement, and impact velocity of projectiles under altered gravitational acceleration (g = -10.0 m/s²) and specific initial conditions.

Problem 1: Fireworks Shell Explosion

The first scenario involves a fireworks shell launched with an initial speed of 100 m/s at a 45° angle. The fuse ignites at the apex of its trajectory, causing the shell to explode. The key questions are: what is the maximum height (explosion height), how long does it take to reach this height, and what is the horizontal displacement by the time of explosion?

Calculations

Initial decomposition of velocity components

The initial velocity components are:

• Horizontal component: v_0x = v₀ cos(45°) = 100 (√2/2) ≈ 70.71 m/s
• Vertical component: v_0y = v₀ sin(45°) = 100 (√2/2) ≈ 70.71 m/s

Maximum height calculation

The apex occurs when vertical velocity becomes zero. Using v_y = v_0y + g*t, and setting v_y = 0:

0 = 70.71 m/s - 10.0 m/s² * t_up

t_up = 70.71 / 10.0 ≈ 7.07 seconds

Height at apex:

h_max = v_0y t_up + 0.5 g * t_up²

h_max = 70.71 7.07 + 0.5 (-10.0) * (7.07)²

h_max ≈ 500.0 - 250.0 = 250.0 meters

Time to reach maximum height is 7.07 seconds, since the ascent time equals the time to reach apex.

Total time of flight:

Since the total flight time is twice the ascent time:

T_total = 2 * 7.07 ≈ 14.14 seconds

Horizontal displacement at explosion:

Horizontal distance:

x = v_0x T_total = 70.71 14.14 ≈ 1000 meters

Problem 2: Volcanic Rock Trajectory

The second scenario involves a volcanic rock ejected with the same initial velocity and launch angle, striking the volcano's side at an altitude 20.0 meters below the starting point. The task is to compute the flight time and impact velocity vector.

Calculations

Establish vertical displacement equation:

Vertical displacement (Δy) = y_final - y_initial = -20.0 m (since it strikes 20 m below launch point)

Vertical kinematic equation:

Δy = v_0y t + 0.5 g * t²

-20.0 = 70.71 t + 0.5 (-10.0) * t²

-20.0 = 70.71 t - 5 t²

Rearranged into standard quadratic form:

5 t² - 70.71 t - 20 = 0

Quadratic formula application:

t = [70.71 ± √(70.71)² - 4 5 (-20)] / (2 * 5)

t = [70.71 ± √(5000 + 400)] / 10

t = [70.71 ± √5400] / 10

t = [70.71 ± 73.48] / 10

Two solutions:

• t₁ = (70.71 + 73.48) / 10 ≈ 14.22 seconds
• t₂ = (70.71 - 73.48) / 10 ≈ -0.27 seconds (discarded as negative time)

Thus, the flight time to impact is approximately 14.22 seconds.

Impact velocity:

Horizontal velocity remains unchanged:

v_x = 70.71 m/s

Vertical velocity at impact:

v_y = v_0y + g t = 70.71 - 10.0 14.22 ≈ 70.71 - 142.2 ≈ -71.49 m/s

The magnitude of impact velocity:

v_impact = √(v_x)² + (v_y)² ≈ √(70.71)² + (-71.49)² ≈ √5000 + 5112 ≈ √10112 ≈ 100.56 m/s

The impact velocity's direction is approximately:

θ = arctangent(|v_y| / v_x) ≈ arctangent(71.49 / 70.71) ≈ 45° below horizontal.

Conclusion

Adjusting initial conditions and gravitational acceleration significantly affects the projectile’s maximum height, flight duration, horizontal displacement, and impact velocity. These calculations demonstrate the importance of kinematic equations in predicting projectile trajectories in both engineered and natural phenomena, such as fireworks displays and volcanic eruptions. The results are consistent with theoretical expectations and serve as valuable tools in physics analysis and natural hazard assessment.

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