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Chapter 4: Problem 91

During volcanic eruptions, chunks of solid rock can be blasted out of the volcano; these projectiles are called volcanic bombs. Figure \(4-51\) shows a cross section of Mt. Fuji, in Japan. (a) At what initial speed would a bomb have to be ejected, at angle \(\theta_{0}=35^{\circ}\) to the horizontal, from the vent at \(A\) in order to fall at the foot of the volcano at \(B\), at vertical distance \(h=\) \(3.30 \mathrm{~km}\) and horizontal distance \(d=9.40 \mathrm{~km}\) ? Ignore, for the moment, the effects of air on the bomb's travel. (b) What would be the time of flight? (c) Would the effect of the air increase or decrease your answer in (a)?

Short Answer

Expert verified
The initial speed required is approximately 172.8 m/s, and the time of flight is about 61.5 s. Air resistance would increase the required initial speed.

Step by step solution

01

Break Down the Problem

We are given the initial angle of projection \( \theta_0 = 35^{\circ} \) and need to find the initial speed \( v_0 \), horizontal distance \( d = 9.40 \text{ km} \), and vertical distance \( h = 3.30 \text{ km} \). We ignore air resistance and use projectile motion equations to determine the initial speed and time of flight.
02

Set Up the Equations of Motion

For projectile motion, the horizontal and vertical displacements \( d \) and \( h \) can be expressed in terms of the initial velocity \( v_0 \):1. Horizontal motion: \[ d = v_0 \cos\theta_0 \cdot t \]2. Vertical motion: \[ h = v_0 \sin\theta_0 \cdot t - \frac{1}{2}gt^2 \]where \( g = 9.81 \text{ m/s}^2 \) is the acceleration due to gravity.
03

Solve for Time of Flight

Solve the horizontal motion equation for \( t \):\[ t = \frac{d}{v_0 \cos\theta_0} \]Substitute for \( t \) in the vertical motion equation to find:\[ h = v_0 \sin\theta_0 \cdot \frac{d}{v_0 \cos\theta_0} - \frac{1}{2}g \left(\frac{d}{v_0 \cos\theta_0}\right)^2 \]This simplifies to:\[ h = d \tan\theta_0 - \frac{gd^2}{2v_0^2 \cos^2\theta_0} \].
04

Solve for Initial Speed

Rearrange for the initial speed \( v_0 \):\[ v_0^2 = \frac{gd^2}{2(d \tan\theta_0 - h) \cos^2\theta_0} \]Substitute given values \( g = 9.81 \text{ m/s}^2 \), \( d = 9400 \text{ m} \), \( h = 3300 \text{ m} \), and \( \theta_0=35^{\circ} \):\[ v_0^2 = \frac{9.81 \times 9400^2}{2(9400 \tan 35^{\circ} - 3300) \cos^2 35^{\circ}} \]After calculating, find \( v_0 \).
05

Calculate Time of Flight

Substitute \( v_0 \) back into:\[ t = \frac{d}{v_0 \cos\theta_0} \]This gives the time of flight \( t \).
06

Consider the Effects of Air Resistance

Air resistance would generally decrease the projectile's range and speed, thus increasing the required initial speed for the volcanic bomb to reach the target location at \( B \).

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Key Concepts

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

Understanding Initial Velocity Calculation
In projectile motion, initial velocity is crucial because it determines how far and how high an object will travel. For volcanic bombs being ejected from a volcano, the initial speed, represented by \(v_0\), must be calculated to understand the bomb's trajectory.

Here's how to approach it:
  • Break down the motion into horizontal and vertical components. The initial velocity \(v_0\) is split into two parts: horizontal \(v_{0_x}\) and vertical \(v_{0_y}\).
  • Use trigonometry: \( v_{0_x} = v_0 \cos\theta_0 \) and \( v_{0_y} = v_0 \sin\theta_0 \). The angle of projection \(\theta_0\) is given as \(35^{\circ}\).
  • In the problem, you need the horizontal \(d = 9.40 \text{ km}\) and vertical distances \(h = 3.30 \text{ km}\).
The formula for calculating \(v_0\) considering these components comes from plugging those distances into the respective projectile motion equations.Rearrange the formula for \(v_0\) using given values:\[ v_0^2 = \frac{gd^2}{2(d \tan\theta_0 - h) \cos^2\theta_0}\]By entering these values into the equation, we solve for \(v_0\), giving the necessary speed to reach point \(B\).
Calculating Time of Flight in Projectile Motion
Time of flight refers to the total time the projectile spends in the air from launch until it hits the ground. It is significant because it directly relates to the distance and speed of the projectile.

Here's how you can calculate it:
  • Initially, solve for \(t\) using the horizontal motion: \[ t = \frac{d}{v_0 \cos\theta_0} \]
  • The given values are substituted into this equation, allowing us to calculate the time based on the horizontal movement.
  • Substituting the initial velocity \(v_0\) obtained from the previous section gives the complete time of flight for the volcanic bomb to travel from \(A\) to \(B\).
Understanding the time of flight helps in predicting where and when the projectile will land, which is particularly useful in safety and evacuation scenarios near volcanic regions.
Effects of Air Resistance on Projectile Motion
Air resistance is a force that acts opposite to the direction of the projectile's motion. Although ignored in simple projectile motion problems, it's an important consideration in real-world scenarios like volcanic eruptions.

Impact on the projectile:
  • Decrease in Range: Air resistance can significantly reduce the horizontal distance a projectile can cover.
  • Reduced Speed: As the projectile moves, air resistance slows it down, requiring a greater initial velocity to maintain the expected range.
  • Altered Trajectory: The path looks more curved and the maximum height is lower than predicted without air resistance.
In this exercise, acknowledging air resistance would mean recalculating \(v_0\) to be higher than the original computation, ensuring the volcanic bomb still reaches point \(B\). This adjustment is necessary for precise predictions and planning, especially in volcanic regions prone to activity. Recognizing these forces provides a more holistic understanding of projectile motion.

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Most popular questions from this chapter

Oasis \(A\) is \(90 \mathrm{~km}\) due west of oasis \(B .\) A desert camel leaves \(A\) and takes \(50 \mathrm{~h}\) to walk \(75 \mathrm{~km}\) at \(37^{\circ}\) north of due east. Next it takes \(35 \mathrm{~h}\) to walk \(65 \mathrm{~km}\) due south. Then it rests for \(5.0 \mathrm{~h}\). What are the (a) magnitude and (b) direction of the camel's displacement relative to \(A\) at the resting point? From the time the camel leaves \(A\) until the end of the rest period, what are the (c) magnitude and (d) direction of its average velocity and (e) its average speed? The camel's last drink was at \(A ;\) it must be at \(B\) no more than \(120 \mathrm{~h}\) later for its next drink. If it is to reach \(B\) just in time, what must be the (f) magnitude and (g) direction of its average velocity after the rest period?

A cat rides a merry-go-round turning with uniform circular motion. At time \(t_{1}=2.00 \mathrm{~s}\), the cat's velocity is \(\vec{v}_{1}=\) \((3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\), measured on a horizontal \(x y\) coordinate system. At \(t_{2}=5.00 \mathrm{~s}\), the cat's velocity is \(\vec{v}_{2}=(-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+\) \((-4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\), What are (a) the magnitude of the cat's centripetal acceleration and (b) the cat's average acceleration during the time interval \(t_{2}-t_{1}\), which is less than one period?

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The minute hand of a wall clock measures \(10 \mathrm{~cm}\) from its tip to the axis about which it rotates. The magnitude and angle of the displacement vector of the tip are to be determined for three time intervals. What are the (a) magnitude and (b) angle from a quarter after the hour to half past, the (c) magnitude and (d) angle for the next half hour, and the (e) magnitude and (f) angle for the hour after that?

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