/*! 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 50 A chinook salmon has a maximum u... [FREE SOLUTION] | 91Ó°ÊÓ

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A chinook salmon has a maximum underwater speed of \(3.58 \mathrm{~m} / \mathrm{s}\), but it can jump out of water with a spced of \(6.26 \mathrm{~m} / \mathrm{s}\). To move upstream past a waterfall, the salmon does not need to jump to the top of the fall. but only to a point in the fall where the water speed is less than \(3.58 \mathrm{~m} / \mathrm{s}\); it can then swim up the fall for the remaining distance. Because the salmon must make forward progress in the water, let's assume it can swim to the top if the water speed is \(3.00 \mathrm{~m} / \mathrm{s}\). If water has a speed of \(1.50 \mathrm{~m} / \mathrm{s}\) as it passes over a ledge, how far below the ledge will the water be moving with a speed of \(9.00 \mathrm{~m} / \mathrm{s}^{2}\) (Note that water undergoes projectile motion once it leaves the ledge.) If the salmon is able to jump vertically upward from the base of the fall, what is the maximum height of waterfall that the salmon can clear?

Short Answer

Expert verified
The maximum height of the waterfall that the salmon can clear is the sum of the distance the water falls till it reaches a speed of 3.00 m/s and the maximum jumping height of the salmon.

Step by step solution

01

Calculation of the water's falling distance till it reaches speed of 3.00 m/s

Firstly, the equation for motion is used to calculate the distance the water falls till it reaches a speed of 3.00 m/s. The equation is given as \(v^2 = u^2 + 2gs\) where \(v\) is final velocity, \(u\) is initial velocity, \(g\) is acceleration due to gravity and \(s\) is the distance. Re-arranging and solving for \(s\), we find that \(s = (v^2 - u^2) / 2g\). Plugging the values \(v = 3.00 m/s\), \(u = 1.50 m/s\), \(g = 9.81 m/s^2\), we get the calculated distance.
02

Calculation of the maximum jumping height of the salmon

Based on the speed the salmon can achieve underwater, we calculate the maximum height it can reach when it jumps out. Here, the maximum height is derived by using the equation of motion \(v^2 = u^2 + 2gs\) again, where this time \(v = 0 m/s\) as the final velocity of the salmon when it reaches the maximum height is 0. The initial velocity \(u\) is the speed of the salmon, which is 6.26 m/s. Solving for \(s\), it is found that \(s = (v^2 - u^2) / 2g\). Substituting the values \(v = 0 m/s\), \(u = 6.26 m/s\), \(g = 9.81 m/s^2\) leads to the maximum height.
03

Calculation of the maximum height of the waterfall that the salmon can clear

Finally, to find the maximum height of the waterfall that the salmon can clear, we sum the distance that the water falls until it reaches a speed of 3.00 m/s and the maximum jumping height of the salmon. Both of these distances are calculated in the previous steps.

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Key Concepts

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

Kinematics Equations
Kinematics equations, also known as the equations of motion, play a central role in understanding the movement of any object under the influence of forces. In the context of projectile motion, they help us track the object's velocity and position over time. For a salmon jumping to clear a waterfall, these equations can be used to identify the maximum height it can reach.

One of the key equations is expressed as \(v^2 = u^2 + 2as\), where \(v\) represents the final velocity of the object, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the displacement. By rearranging and solving these kinematic equations, we can determine the distance an object travels or the height it reaches during its motion.

To ensure that students fully grasp the concept, it's crucial to emphasize that the initial velocity plays a significant part in the calculations. For instance, if the salmon has an initial upward speed, this directly influences how high it can jump. Students should be encouraged to visualize the scenario, break down the motion into horizontal and vertical components, and apply the kinematics equation accordingly.
Acceleration Due to Gravity
The acceleration due to gravity, denoted \(g\), is approximately \(9.81 \, \text{m/s}^2\) on the surface of Earth. It's the rate at which an object accelerates towards Earth when it's in free fall, meaning that it affects the motion of projectiles like water in the waterfall scenario.

In physics problems, like the ones involving salmon jumping, \(g\) is a constant that provides the acceleration in the kinematics equations. Understanding that this acceleration acts downward is central to solving projectile motion problems. The direction of gravity pulls objects back towards the Earth, which is why the salmon, after propelling itself upward, eventually comes back down;

When giving exercises improvement advice, remind students to always incorporate \(g\) as a negative value when solving for upward motion since it opposes the initial upward velocity. This helps in accurately calculating the maximum height of objects like the jumping salmon.
Motion of Projectiles
Projectile motion refers to the movement of an object thrown or projected into the air, subject only to acceleration due to gravity. This concept is key in our salmon example as it ties together the initial horizontal velocity and the vertical force of gravity impacting the salmon's jump.

A projectile typically moves along a curved path known as a parabola. This motion can be broken down into two components: horizontal motion, which is constant because it's unaffected by gravity, and vertical motion, which is influenced by gravity and changes over time. For students to effectively understand projectile motion, it's beneficial to analyze the separate motions along the x-axis (horizontal) and y-axis (vertical).

Remaining focused on important aspects like the initial speed of the projectile, the angle of projection, and the effects of gravity will enable students to conceptualize how these factors influence the trajectory. In doing so, they will better understand the comprehensive motion of projectiles, aiding in the solution of complex problems like calculating the height of a waterfall a salmon can clear.

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Most popular questions from this chapter

A river has a steady speed of \(v_{1}\) A student swims upstream a distance \(d\) and back to the starting point. (a) If the student can swim at a speed of \(v\) in still water, how much time \(t_{\text {op }}\) does it take the student to swim upstream a distance \(d ?\) Express the answer in terms of \(d\). \(v_{\text {, }}\) and \(v_{s}\). (b) Using the same variables, how much time \(t_{\text {doun }}\) does it take to swim back downstream to the starting point? (c) Sum the answers found in parts (a) and (b) and show that the time \(t_{a}\) required for the whole trip can be written as $$t_{\mathrm{a}}=\frac{2 d / v}{\mathrm{I}-\mathrm{v}_{1}^{2} / v^{2}}$$ (d) How much time \(t_{b}\) does the trip take in still water? (e) Which is larger, \(l_{a}\) or \(t_{b}^{2}\) Is it always larger?

A river flows due east at \(1.50 \mathrm{~m} / \mathrm{s}\). A boat crosses the river from the south shore to the north shore by maintaining a constant velocity of \(10.0 \mathrm{~m} / \mathrm{s}\) due north relative to the water. (a) What is the velocity of the boat relative to the shore? (b) If the river is \(300 \mathrm{~m}\) wide, how far downstream has the boat moved by the time it reaches the north shore?

The determined Wile \(\mathrm{E}\). Coyote is out once more to try to capture the elusive roadrunner. The coyote wears a new pair of Acme power roller skates, which provide a constant horizontal acceleration of \(15 \mathrm{~m} / \mathrm{s}^{2}\), as shown in Figure Pg.73. The coyote starts off at rest \(70 \mathrm{~m}\) from the cdge of a cliff at the instant the roadrunner zips by in the direction of the cliff. (a) If the roadrunner moves with constant speed, find the minimum speed the roadrunner must have to reach the cliff before the coyote. (b) If the cliff is \(100 \mathrm{~m}\) above the base of a canyon, find where the coyote lands in the canyon. (Assume his skates are still in operation when he is in "flight" and that his horizontal component of acceleration remains constant at \(15 \mathrm{~m} / \mathrm{s}^{2} .\) )

A jet airliner moving initially at \(3.00 \times 10^{2} \mathrm{mi} / \mathrm{h}\) due east enters a region where the wind is blowing \(1.00\) \(\times 10^{2} \mathrm{mi} / \mathrm{h}\) in a direction \(30.0^{\circ}\) north of east. (a) Find the components of the velocity of the jet airliner relative to the air, \(\vec{v}_{\mu}\). (b) Find the components of the velocity of the air relative to Earth, \(\overrightarrow{\mathbf{v}}_{M \mathcal{E}}\). (c) Write an equation analo- gous to Equation \(3.16\) for the velocities \(\vec{v}_{j}, \vec{v}_{A B}\), and \(\vec{v}_{g E}\) (d) What is the speed and direction of the aircraft relative to the ground?

A quarterback throws a football toward a receiver with an initial speed of \(20 \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ}\) above the horizontal. At that instant the receiver is \(20 \mathrm{~m}\) from the quarterback. In what direction and with what constant speed should the receiver run in order to catch the football at the level at which it was thrown?

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