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Chapter 4: Problem 75

A battleship staring at \(13 \mathrm{~m} / \mathrm{s}\) toward the shore fires a shot. in the forward direction. The elevation angle of the gun is \(20^{\circ}\), and the mtrale speed of the shot is \(660 \mathrm{~m} / \mathrm{s}\). What is the velocity vector of the shot relative to the shore?

Short Answer

Expert verified
The velocity vector of the shot relative to the shore is \((632.34 \mathrm{~m/s}, 225.24 \mathrm{~m/s})\).

Step by step solution

01

Determine Components of Muzzle Velocity

The shot's initial velocity has two components: horizontal and vertical. Use trigonometry to find them. The muzzle speed is given as \(v_m = 660 \mathrm{~m/s}\), with an elevation angle \(\theta = 20^{\circ}\).For the horizontal component: \[v_{m_x} = v_m \cos(\theta) = 660 \cdot \cos(20^{\circ})\]For the vertical component:\[v_{m_y} = v_m \sin(\theta) = 660 \cdot \sin(20^{\circ})\]
02

Calculate Horizontal Component Relative to Shore

The battleship has a velocity of \(13 \mathrm{~m/s}\) toward the shore, adding to the horizontal component of the shot. Thus, the horizontal component of the shot's velocity relative to the shore is:\[v_{relative_x} = v_{m_x} + 13 \mathrm{~m/s}\]
03

Maintain Vertical Component Relative to Shore

The vertical component is unaffected by the movement of the battleship since it's perpendicular to its direction. Therefore, the vertical component of the velocity relative to the shore is simply:\[v_{relative_y} = v_{m_y}\]
04

Determine Velocity Vector

The shot's velocity relative to the shore is a combination of its horizontal and vertical components. Therefore, the velocity vector \(\mathbf{v}_{relative}\) can be expressed as:\[\mathbf{v}_{relative} = (v_{relative_x}, v_{relative_y})\]
05

Final Calculation

Calculate the numerical results for the components:\[v_{m_x} = 660 \cdot \cos(20^{\circ}) \approx 619.34 \mathrm{~m/s}\]\[v_{m_y} = 660 \cdot \sin(20^{\circ}) \approx 225.24 \mathrm{~m/s}\]Thus, the relative horizontal component becomes:\[v_{relative_x} = 619.34 + 13 = 632.34 \mathrm{~m/s}\]The vertical component:\[v_{relative_y} = 225.24 \mathrm{~m/s}\]Therefore, the velocity vector relative to the shore is:\[\mathbf{v}_{relative} = (632.34 \mathrm{~m/s}, 225.24 \mathrm{~m/s})\]

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Key Concepts

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

Velocity Vector
In the context of projectile motion, a velocity vector describes the speed and direction of an object. For this problem, we need to understand the velocity vector of a projectile shot from a moving battleship. The velocity vector is composed of two components: horizontal and vertical.
  • The horizontal component accounts for the forward motion of the shot; it combines the speed at which the battleship moves and the speed imparted by the gun.
  • The vertical component remains independent of the battleship's motion as it represents the projectile's upwards velocity.
To find the overall velocity vector, we combine these two components mathematically. This results in a vector with specific magnitudes in both directions, giving us the full description of the shot's movement relative to the shore.
Trigonometry in Physics
Trigonometry is a powerful tool used in physics to resolve vectors into their components, especially in projectile motion problems. With trigonometry, you can take an angled vector and break it down into its horizontal and vertical parts.
In this problem, we use a given muzzle velocity of the shot fired at an angle. The angle helps us determine the impacts of gravity and forward motion:
  • The cosine of the angle gives us the proportion of the velocity in the horizontal direction:
    \[v_{m_x} = v_m \cos(\theta)\]
  • The sine of the angle gives us the proportion of the velocity in the vertical direction:
    \[v_{m_y} = v_m \sin(\theta)\]
These calculations are central to accurately assessing the motion of projectiles in physics, allowing us to predict their paths and behaviors.
Relative Motion
Relative motion is a concept that explains how the velocity of an object depends upon the observer's point of view. In this problem, the shot fired from the battleship has different velocity readings: one from the battleship's perspective and another from the shore's perspective.
The battleship itself moves toward the shore, adding to the horizontal velocity of the shot from the gun. This needs to be accounted for to find the shot's true horizontal speed relative to the shore.
  • The horizontal velocity of the battleship is added to the shot's horizontal velocity calculated via trigonometry to find the total relative horizontal velocity viewed from the shore.
  • The vertical velocity component remains unchanged, as it is perpendicular to the direction of the battleship's movement.
Understanding relative motion helps us grasp how velocities merge and impact an observer's perception, which is vital in various fields, including navigation and space travel.

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Most popular questions from this chapter

The blade of a circular saw has a diarneter of \(20 \mathrm{~cm}\). If this blade fotatee at 7000 revolutions per minute (its maximum safe ipeed), what arc the spced and the centriperal acccicration of a point on the rim?

A hattelebip steming at \(45 \mathrm{~km} / \mathrm{h}\) fires u gun at right angles to the longitudinal axis of the ship. The cicvation angle of the gun is \(30^{\prime}\), and the muzde velncity of the shot is \(720 \mathrm{~m} / \mathrm{s}\) the gravitational acccieration in \(9.8 \mathrm{~m} / \mathrm{s}^{2}\), What is the range of thin abot in therefcrence frame of the grocnd? Prctend that there is no air retistaroc.

At the cntrance to Ambrone Channel at New York hubor, thc maximum tidal carrent has a velocity of \(4.2 \mathrm{~km} / \mathrm{h}\) in a dircetion \(20^{\circ}\) south of eact. What is the component of this velocity] in the east direction? In the north direction?

'Two foothall players are initially \(15 \mathrm{~m}\) apart. The firot player (a rerciver) runn perpendicular to the line joining the two players at a contant ipeed of \(8.0 \mathrm{~m} / \mathrm{s}\). Affer two ticconds, the wecond player (the quarterback) thrims the hall with a horimontal omponcnt vclocity of \(20 \mathrm{~m} / \mathrm{s}\). In what hotizontal directicn and with what vertical launch ungle should the quarterback throw so that the hull reaches the same -pot the receivcr will be? At what time will the ball be caught?

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