/*! 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 12 A daring \(510-N\) swimmer dives... [FREE SOLUTION] | 91Ó°ÊÓ

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A daring \(510-N\) swimmer dives off a cliff with a running horizontal leap, as shown in Fig. 3.39 . What must her mini- mum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom. which is 1.75 \(\mathrm{m}\) wide and 9.00 \(\mathrm{m}\) below the top of the cliff?

Short Answer

Expert verified
The minimum speed needed is 1.29 m/s.

Step by step solution

01

Understanding the Problem

The swimmer dives horizontally off a cliff. The goal is to find the minimum horizontal velocity needed to avoid a ledge that is 1.75 meters wide, 9.00 meters below the top of the cliff.
02

Analyze the Vertical Motion

Use the formula for free fall to determine the time taken to fall 9.00 meters vertically. The equation is \( h = \frac{1}{2} g t^2 \), where \( h = 9.00 \text{ m} \) and \( g = 9.81 \text{ m/s}^2 \). Calculating gives \( 9 = 0.5 \times 9.81 \times t^2 \), solving for \( t \), we get \( t^2 = \frac{18}{9.81} \approx 1.834 \), hence \( t \approx \sqrt{1.834} \approx 1.354 \text{ s} \).
03

Analyze the Horizontal Motion

The horizontal motion is uniform, meaning the horizontal velocity is constant. We need to find the minimum horizontal velocity to cover a 1.75-meter ledge in the time found in Step 2. Use the formula \( x = v_x t \), where \( x = 1.75 \text{ m} \), \( t = 1.354 \text{ s} \). Substitute to find the minimum speed \( v_x = \frac{1.75}{1.354} \approx 1.29 \text{ m/s} \).
04

Conclusion

To ensure the swimmer clears the ledge, the calculated minimum horizontal velocity required as she leaves the cliff is approximately 1.29 m/s. This speed allows her to cover the horizontal distance in the duration she is in free fall.

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Key Concepts

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

Free Fall
In the realm of physics, free fall is a specific type of motion where an object experiences acceleration due to gravity alone. When an object is in free fall, it does not encounter any other forces, such as air resistance. The primary force acting upon it is the pull of gravity, resulting in a vertical acceleration directed downwards.

For earth-bound objects, the acceleration due to gravity is approximately 9.81 m/s². Using this constant, we can predict how quickly an object will fall over time, which is crucial for solving problems like the one in the exercise. In our scenario, the swimmer falls 9 meters vertically, and we use the equation for free fall: \[ h = \frac{1}{2} g t^2 \]where:
  • h is the height (9.00 m in this case)

  • g is the acceleration due to gravity (9.81 m/s²)
  • t is the time in seconds
Rearranging the equation to solve for time (t), we calculate how long the swimmer is in the air before potentially hitting the ledge. The process involves substituting the known values and solving for t, leading to a time of approximately 1.354 seconds.
Horizontal Velocity
Horizontal velocity refers to the constant speed at which an object travels horizontally. In projectile motion, while vertical velocity changes due to gravity, the horizontal component remains steady, assuming negligible air resistance. This distinction is important when analyzing a horizontally launched projectile, like the swimmer in the exercise.

To determine the minimum horizontal velocity, one must ensure that the swimmer covers the horizontal distance (1.75 m) in the time calculated for the free fall. This calculation uses the formula:\[ x = v_x t \]
  • x represents the horizontal distance to be cleared (1.75 m)

  • v_x is the horizontal velocity
  • t is the time in seconds (1.354 s)
Using this relation, we find that the swimmer needs a horizontal velocity of approximately 1.29 m/s to clear the ledge. This speed effectively stretches the horizontal travel over the time descended, ensuring a safe leap from the cliff.
Kinematic Equations
Kinematic equations are pivotal in solving problems involving motion, especially when detailing projectile paths. They provide mathematical tools to describe an object's motion using variables such as initial velocity, final velocity, time, and acceleration. The primary kinematic equation used here is:\[ d = v_i t + \frac{1}{2} a t^2 \]
  • d stands for the distance covered

  • v_i is the initial velocity
  • a represents acceleration (gravity for vertical motion)
  • t is the time under observation
In the current exercise, these equations allow for the separate analysis of vertical (free fall) and horizontal (constant velocity) motions. The separation simplifies solving complex motion because it acknowledges that horizontal and vertical components of motion don’t influence each other in a projectile. This approach streamlines calculating how the swimmer clears the ledge. It underscores that in projectiles, one must consider two motions together to solve the problem effectively.

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Most popular questions from this chapter

Spiraling Up. It is common to see birds of prey rising upward on thermals. The paths they take may be spiral-like. You can model the spiral motion as uniform circular motion combined with a constant upward velocity. Assume a bird completes a circle of radius 8.00 \(\mathrm{m}\) every 5.00 \(\mathrm{s}\) and rises vertically at a rate of 3.00 \(\mathrm{m} / \mathrm{s}\) . Determine: (a) the speed of the bird relative to the ground; \((b)\) the bird's acceleration (magnitude and direction); and (c) the angle between the bird's velocity vector and the horizontal.

The earth has a radius of 6380 \(\mathrm{km}\) its axis in 24 \(\mathrm{h}\) . (a) What is the radial acceleration of an object at the earth's equator? Give your answer in \(\mathrm{m} / \mathrm{s}^{2}\) and as a fraction of g. (b) If \(a_{n d}\) at the equator is greater than \(g\) , objects would fly off the earth's surface and into space. (We will see the reason for this in Chapter \(5 . .\) What would the period of the earth's rotation have to be for this to occur?

Two tanks are engaged in a training exercise on level ground. The first tank fires a paint-filled training round with a muzzle speed of 250 \(\mathrm{m} / \mathrm{s}\) at \(10.0^{\circ}\) above the horizontal while advancing toward the second tank with a speed of 15.0 \(\mathrm{m} / \mathrm{s}\) relative to the ground. The second tank is retreating at 35.0 \(\mathrm{m} / \mathrm{s}\) relative to the ground, but is hit by the shell. You can ignore air resistance and assume the shell hits at the same height above ground from which it was fired. Find the distance between the tanks (a) when the round was first fired and (b) at the time of impact.

A squirrel has \(x-\) and \(y\) -coordinates \((1.1 \mathrm{m}, 3.4 \mathrm{m})\) at time \(t_{1}=0\) and coordinates \((5.3 \mathrm{m},-0.5 \mathrm{m})\) at time \(t_{2}=3.0 \mathrm{s} .\) For this time interval, find (a) the components of the average velocity, and \((\mathrm{b})\) the magnitude and direction of the average velocity.

A 400.0 -m-wide river flows from west to east at 30.0 \(\mathrm{m} / \mathrm{min}\) . Your boat moves at 100.0 \(\mathrm{m} / \mathrm{min}\) relative to the water no matter which direction you point it. To cross this river, you start from a dock at point \(A\) on the south bank. There is a boat landing directly opposite at point \(B\) on the north bank, and also one at point \(C, 75.0 \mathrm{m}\) downstream from \(B(\text { Fig. } 3.53)\) . (a) Where on the north shore will you land if you point your boat perpendicular to the water current, and what distance will you have traveled? (b) If you initially aim your boat directly toward point \(C\) and do not change that bearing relative to the shore, where on the north shore will you land? (c) To reach point \(C :(i)\) at what bearing must you aim your boat, (ii) how long will it take to cross the river, (iii) what distance do you travel, and (iv) and what is the speed of your boat as measured by an observer standing on the river bank?

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