/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 The small archerfish (length 20 ... [FREE SOLUTION] | 91Ó°ÊÓ

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The small archerfish (length 20 to \(25 \mathrm{cm}\) ) lives in brackish waters of southeast Asia from India to the Philippines. This aptly named creature captures its prey by shooting a stream of water drops at an insect, either flying or at rest. The bug falls into the water and the fish gobbles it up. The archerfish has high accuracy at distances of \(1.2 \mathrm{m}\) to \(1.5 \mathrm{m}\) and it sometimes makes hits at distances up to 3.5 m. A groove in the roof of its mouth, along with a curled tongue, forms a tube that enables the fish to impart high velocity to the water in its mouth when it suddenly closes its gill flaps. Suppose the archerfish shoots at a target \(2.00 \mathrm{m}\) away, at an angle of \(30.0^{\circ}\) above the horizontal. With what velocity must the water stream be launched if it is not to drop more than \(3.00 \mathrm{cm}\) vertically on its path to the target?

Short Answer

Expert verified
The initial velocity, \(v\), required for the stream of water to reach the target without dropping more than 3.00 cm is found by solving the above equation.

Step by step solution

01

Use the Trigonometric Relationships

We will use trigonometry to find the total time taken. The horizontal component of the velocity, \(v_x\), can be found using cosine of the angle. So, \(v_x = v \cos(30.0^{\circ})\). Since the horizontal distance covered is 2.00 m, the time taken, \(t\), to reach the target will be \(t = 2.00 \, \mathrm{m} / v_x = 2.00 \, \mathrm{m} / (v \cos(30.0^{\circ}))\).
02

Set Up the Equation for Vertical Displacement

We will calculate the vertical displacement, applying the equations of projectile motion. We know that the total drop is 3.00 cm, but the initial upward trajectory due to the launch angle needs to be taken into consideration. So, we set the equation for vertical displacement as follows: \(0.03 \, \mathrm{m} = v \sin(30.0^{\circ}) \cdot t - (1/2) \cdot g \cdot t^2\), where \(g\) is the acceleration due to gravity.
03

Solve for Initial Velocity

Now we substitute \(t\) and \(g\) into the equation. The acceleration due to gravity, \(g\), is approximated as \(9.80 \, \mathrm{m/s^2}\), so we have \(0.03 \, \mathrm{m} = v \sin(30.0^{\circ}) \cdot 2.00 \, \mathrm{m} / (v \cos(30.0^{\circ})) - 0.5 \cdot 9.80 \, \mathrm{m/s^2} \cdot (2.00 \, \mathrm{m} / v \cos(30.0^{\circ}))^2 \). This equation is then solved for the initial velocity, \(v\).

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Key Concepts

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

Understanding Trigonometric Relationships
When studying projectile motion, we employ trigonometric relationships to decompose the motion into horizontal and vertical components. This is crucial because it allows us to analyze the motion in two dimensions separately. For example, if an object is launched at an angle, we can use the sine and cosine functions to determine the initial velocities in the x (horizontal) and y (vertical) directions.

Specifically, if an object is launched with an initial velocity (\(v\)) at an angle (\(\theta\)), the horizontal component of the velocity (\(v_x\)) is given by \(v_x = v \cos(\theta)\), while the vertical component (\(v_y\)) can be found using \(v_y = v \sin(\theta)\). By understanding these relationships, we can proceed to apply the equations of motion to each component independently, which simplifies the analysis of projectile motion.

Using trigonometry this way transforms the problem into a more manageable form and is the first step to finding out how the archerfish adeptly calculates the angle and velocity to capture its prey with a precise jet of water.
Equations of Projectile Motion
Equations of projectile motion come into play after breaking down the velocity into two components through trigonometry. These equations allow us to predict where and when the projectile will arrive at a certain point. In the context of horizontal motion, the distance traveled is given by the horizontal velocity multiplied by time, represented as \(d = v_xt\).

For vertical motion, the equation considers both the vertical component of the initial velocity and the acceleration due to gravity. The vertical displacement \(y\) from the initial point is found using the formula \(y = v_yt - \frac{1}{2}gt^2\), where \(v_y\) is the initial vertical velocity, \(t\) is the time, and \(g\) is the acceleration due to gravity. This equation perfectly models the vertical drop of the water stream shot by the archerfish and helps to calculate the precise velocity needed for a direct hit.

This incorporation of gravitational influence is vital in predicting projectile motion because regardless of the initial velocity, gravity affects all objects uniformly, pulling them downwards at a consistent rate, thus shaping their trajectory.
Acceleration Due to Gravity
The acceleration due to gravity, usually denoted by \(g\), is the rate at which objects accelerate towards the Earth when in free fall. It's a constant value, approximately \(9.80\, \mathrm{m/s^2}\), and plays a key role in the equations of projectile motion. This acceleration is independent of the mass, size, or composition of the falling object, which is why all projectiles, like the archerfish's water stream, experience the same gravitational pull.

When calculating projectile motion, \(g\) affects the vertical component of the motion significantly. It is the reason a projectile eventually falls to the ground after being projected upwards or outwards. For instance, the archerfish must shoot its jet of water with enough initial velocity and at the correct angle to overcome this acceleration and hit the target without the stream falling more than 3 cm.

In the exercise, the influence of gravity is incorporated by adding the term \(\frac{1}{2}gt^2\) to represent the distance fallen due to gravity over time, exemplifying how the acceleration due to gravity is indispensable in the precise calculation of projectile motion.

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Most popular questions from this chapter

Your grandfather is copilot of a bomber, flying horizontally over level terrain, with a speed of 275 m/s relative to the ground, at an altitude of 3000 m. (a) The bombardier releases one bomb. How far will it travel horizontally between its release and its impact on the ground? Neglect the effects of air resistance. (b) Firing from the people on the ground suddenly incapacitates the bombardier before he can call, "Bombs away!" Consequently, the pilot maintains the plane's original course, altitude, and speed through a storm of flak. Where will the plane be when the bomb hits the ground? (c) The plane has a telescopic bomb sight set so that the bomb hits the target seen in the sight at the time of release. At what angle from the vertical was the bomb sight set?

In the What If? section of Example 4.7 , it was claimed that the maximum range of a ski-jumper occurs for a launch angle \(\theta\) given by $$\theta=45^{\circ}-\frac{\phi}{2}$$ where \(\phi\) is the angle that the hill makes with the horizontal in Figure \(4.16 .\) Prove this claim by deriving the equation above.

A high-powered rifle fires a bullet with a muzzle speed of \(1.00 \mathrm{km} / \mathrm{s} .\) The gun is pointed horizontally at a large bull's eye target-a set of concentric rings- \(200 \mathrm{m}\) away. (a) How far below the extended axis of the rifle barrel does a bullet hit the target? The rifle is equipped with a telescopic sight. It is "sighted in" by adjusting the axis of the telescope so that it points precisely at the location where the bullet hits the target at \(200 \mathrm{m}\). (b) Find the angle between the telescope axis and the rifle barrel axis. When shooting at a target at a distance other than \(200 \mathrm{m}\) the marksman uses the telescopic sight, placing its crosshairs to "aim high" or "aim low" to compensate for the different range. Should she aim high or low, and approximately how far from the bull's eye, when the target is at a distance of (c) \(50.0 \mathrm{m},\) (d) \(150 \mathrm{m},\) or (e) \(250 \mathrm{m} ?\) Note: The trajectory of the bullet is everywhere so nearly horizontal that it is a good approximation to model the bullet as fired horizontally in each case. What if the target is uphill or downhill? (f) Suppose the target is \(200 \mathrm{m}\) away, but the sight line to the target is above the horizontal by \(30^{\circ} .\) Should the marksman aim high, low, or right on? (g) Suppose the target is downhill by \(30^{\circ} .\) Should the marksman aim high, low, or right on? Explain your answers.

A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of \(8.00 \mathrm{m} / \mathrm{s}\) at an angle of \(20.0^{\circ}\) below the horizontal. It strikes the ground \(3.00 \mathrm{s}\) later. (a) How far horizontally from the base of the building does the ball strike the ground? (b) Find the height from which the ball was thrown. (c) How long does it take the ball to reach a point \(10.0 \mathrm{m}\) below the level of launching?

A projectile is fired in such a way that its horizontal range is equal to three times its maximum height. What is the angle of projection?

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