Sci212e Physics For This Homework Carefully Study Examples 3

Sci212e Physicsfor This Homework Carefully Study Examples 34 And 35

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, as illustrated in Figure 3.38. 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?

Kilauea in Hawaii is the world's most continuously active volcano. 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, as shown in Figure 3.39. 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? Use g = -10.0 m/s² for gravitational acceleration. Please show all work, not just the answers.

Paper For Above instruction

Projectile motion is a fundamental topic in physics that encompasses the analysis of objects launched into the air under the influence of gravity. Understanding the dynamics of such objects requires the application of kinematic equations, vector components, and trigonometric functions to determine various parameters such as maximum height, time of flight, and horizontal range. This paper delves into two illustrative scenarios of projectile motion, emphasizing the calculations based on an initial velocity of 100 m/s at an angle of 45°, with a gravitational acceleration of -10.0 m/s², slightly adjusted from the typical -9.8 m/s² to explore how variations in parameters influence the outcomes.

First, considering the fireworks shell problem, the key is to determine the height at which the shell explodes. Since the fuse ignites at the highest point, it is necessary to find the time to reach this point, which occurs at the peak of the trajectory. The vertical component of the initial velocity, v0y, is given by v0 sin(45°) = 100 (√2 / 2) ≈ 70.71 m/s. At the maximum height, vertical velocity v_y equals zero, and using the kinematic equation v_y = v0y + g * t, solving for the time to reach the apex yields t_apex = v0y / |g| = 70.71 / 10 ≈ 7.07 seconds.

Next, to find the maximum height, h_max, we employ the equation h_max = v0y t_apex + (1/2) g t_apex². Substituting the known values gives h_max = 70.71 7.07 + 0.5 (-10) (7.07)² ≈ 499.83 - 249.99 ≈ 249.84 meters. Therefore, the shell explodes at approximately 250 meters above the ground.

The total time of flight, T_total, can be computed by recognizing that the total time to ascend and descend are equal for symmetric projectile motion, thus T_total = 2 * t_apex ≈ 14.14 seconds. This duration reflects the total time since launch until the shell hits the ground if it lands back at the initial elevation.

For horizontal displacement, the initial horizontal component of velocity v0x is v0 cos(45°) = 100 (√2 / 2) ≈ 70.71 m/s. The horizontal range at the moment of explosion is then R = v0x t_apex = 70.71 7.07 ≈ 500 meters. However, since the fuse ignites at the peak, the horizontal distance traveled until explosion corresponds to this value, approximately 500 meters.

In the second scenario, involving the ejection of a large rock from Kilauea volcano, the problem asks for the time taken for the rock to reach a point 20 meters below its starting elevation. The initial vertical and horizontal velocity components mirror the previous case: v0x ≈ 70.71 m/s and v0y ≈ 70.71 m/s.

The vertical motion is governed by the equation y = v0y t + (1/2) g t², where y is the vertical displacement. Since the rock strikes 20 meters below its original height, the displacement y is -20 m. Setting up the equation: -20 = 70.71 t + 0.5 (-10) t² or -20 = 70.71 t - 5 t². Rearranged as 5 t² - 70.71 t - 20 = 0, this quadratic can be solved for t using the quadratic formula t = [70.71 ± √(70.71² - 4 5 (-20))] / (2 * 5).

Calculating the discriminant: 70.71² + 4 5 20 = 5000 + 400 = 5400. Taking the square root yields √5400 ≈ 73.48. Therefore, t = [70.71 ± 73.48] / 10. The positive root gives t = (70.71 + 73.48) / 10 ≈ 144.19 / 10 ≈ 14.42 seconds, which is physically meaningful. The negative root would yield a negative time and is discarded.

To find the velocity at impact, we analyze the components at that instant. The horizontal component remains constant at v_x ≈ 70.71 m/s, whereas the vertical component is v_y = v0y + g t = 70.71 - 10 14.42 ≈ 70.71 - 144.2 ≈ -73.49 m/s, indicating downward velocity at impact. The magnitude of the velocity vector is |v| = √(v_x² + v_y²) ≈ √(70.71² + 73.49²) ≈ √(5000 + 5399) ≈ √10,399 ≈ 101.99 m/s. The direction of this velocity is given by θ = arctangent(v_y / v_x) ≈ arctangent(-73.49 / 70.71) ≈ -45.4°, indicating downward and slightly angled relative to the horizontal.

These calculations demonstrate the application of projectile motion principles, including components analysis, quadratic equations, and kinematics, under specified initial conditions and gravitational acceleration variations. Accurate computation of these parameters is essential in fields such as ballistics, aerospace engineering, and natural phenomena analysis, where trajectory predictions impact safety and operational efficiency.

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