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Chapter 3: Problem 10

A daring 510 -N swimmer dives off a cliff with a running horizontal leap, as shown in Fig. E3.10. What must her minimum 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 must be approximately 1.30 m/s.

Step by step solution

01

Understand the Problem

The swimmer makes a horizontal leap from a cliff. We need to find the minimum speed required for her to clear a ledge that is 1.75 m wide and 9.00 m below her start point. There is no initial vertical velocity since the leap is horizontal.
02

Calculate Time to Fall 9.00 m

We use the equation of motion for vertical displacement since there is acceleration due to gravity. Using the formula \( y = \frac{1}{2}gt^2 \), we can find the time \( t \) it takes to fall 9.00 m. Here, \( y = 9.00 \, \text{m} \) and \( g = 9.81 \, \text{m/s}^2 \). Solving \( 9.00 = \frac{1}{2} \times 9.81 \times t^2 \) gives \( t = \sqrt{\frac{18}{9.81}} \approx 1.35 \text{ s} \).
03

Calculate Minimum Horizontal Speed

With the time \( t = 1.35 \text{ s} \) calculated, use the horizontal motion equation \( x = v_{x}t \) to find the minimum speed \( v_{x} \). Here, \( x = 1.75 \text{ m} \). Solving \( 1.75 = v_{x} \times 1.35 \) results in \( v_{x} \approx 1.30 \text{ m/s} \).

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Key Concepts

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

Understanding Horizontal Leap
In projectile motion, a horizontal leap refers to the movement of an object in the horizontal direction, without any initial vertical velocity. This type of motion is important when considering athletes jumping across gaps or, in this context, a swimmer diving off a cliff.
In our exercise, the swimmer's horizontal speed at take-off is crucial to ensure she successfully clears the ledge at the bottom. Since she leaps horizontally, she starts with only horizontal velocity, which means:
  • Her vertical velocity at the start is zero.
  • The horizontal velocity remains constant throughout the flight, as there are no forces acting in the horizontal direction (assuming no air resistance).
Understanding horizontal leap helps simplify calculations. We only need to focus on maintaining a sufficient speed to cover the gap before hitting the ground.
Exploring Vertical Displacement
Vertical displacement in projectile motion refers to the distance an object falls due to gravity. It is important because it allows the calculation of time in flight.
The swimmer dives from a height of 9.00 m, and her fall can be described by the vertical motion equation:
  • \[ y = \frac{1}{2}gt^2 \]
Here, \(y\) is the vertical displacement (9.00 m), and \(g\) is the acceleration due to gravity (9.81 m/s虏).
With no initial vertical velocity, this equation helps determine the time it takes for her to fall from the top to the bottom of the cliff, which we found to be 1.35 seconds. This time is essential for calculating how far she can travel horizontally.
Understanding Gravity Acceleration
Gravity acceleration plays a crucial role in projectile motion. By acting on objects, it causes them to accelerate downward, influencing their vertical motion.
For any object near the Earth's surface, gravity provides a constant acceleration of approximately 9.81 m/s虏 downward. This acceleration affects only the vertical component of motion, while the horizontal component remains unaffected.
  • In the scenario of the swimmer, it determines how fast she'll fall downwards.
  • The time it takes to fall a certain height can be calculated using gravity acceleration.
Understanding gravity's effect helps predict where and when an object will land, crucial for ensuring clearance over obstacles. By using gravity, we can determine how long it takes to reach the ground and plan the necessary horizontal speed.
Applying Motion Equations
Motion equations are tools that help us describe the movement of objects. In projectile motion, we frequently apply these equations separately to horizontal and vertical movements.
For our cliff diving exercise, two main equations were used:
  • Vertical Motion: \[ y = \frac{1}{2}gt^2 \] calculates the time of fall.
  • Horizontal Motion: \[ x = v_{x}t \] determines the necessary horizontal speed to cover a given distance.
By determining the time of fall, we apply it in the horizontal equation to find how fast the swimmer needs to leap. The calculated speed, 1.30 m/s, ensures she'll clear the ledge. Splitting the motion into these two components and applying the respective equations is a simple yet effective way to solve these problems.

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Most popular questions from this chapter

When a train's velocity is 12.0 \(\mathrm{m} / \mathrm{s}\) eastward, raindrops that are falling vertically with respect to the earth make traces that are inclined \(30.0^{\circ}\) to the vertical on the windows of the train. (a) What is the horizontal component of a drop's velocity with respect to the earth? With respect to the train? (b) What is the magnitude of the velocity of the raindrop with respect to the earth? With respect to the train?

A student sits atop a platform a distance \(h\) above the ground. He throws a large firecracker horizontally with a speed \(v .\) However, a wind blowing parallel to the ground gives the firecracker a constant horizontal acceleration with magnitude a. This results in the firecracker reaching the ground directly under the student. Determine the height \(h\) in terms of \(v, a,\) and \(g .\) You can ignore the effect of air resistance on the vertical motion.

In an action-adventure film, the hero is supposed to throw a grenade from his car, which is going \(90.0 \mathrm{km} / \mathrm{h},\) to his enemy's car, which is going 110 \(\mathrm{km} / \mathrm{h}\) . The enemy's car is 15.8 \(\mathrm{m}\) in front of the hero's when he lets go of the grenade. If the hero throws the grenade so its initial velocity relative to him is at an angle of \(45^{\circ}\) above the horizontal, what should the magnitude of the initial velocity be? The cars are both traveling in the same direction on a level road. You can ignore air resistance. Find the magnitude of the velocity both relative to the hero and relative to the earth.

A test rocket is launched by accelerating it along a 200.0 -m incline at 1.25 \(\mathrm{m} / \mathrm{s}^{2}\) starting from rest at point \(A\) (Fig. P3.47). The incline rises at \(35.0^{\circ}\) above the horizontal, and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored). Find (a) the maximum height above the ground that the rocket reaches, and (b) the greatest horizontal range of the rocket beyond point \(A\) .

In U.S. football, after a touchdown the team has the opportunity to earn one more point by kicking the ball over the bar between the goal posts. The bar is 10.0 \(\mathrm{ft}\) above the ground, and the ball is kicked from ground level, 36.0 \(\mathrm{ft}\) horizontally from the bar (Fig. P3.62). Football regulations are stated in English units, but convert them to SI units for this problem.(a) There is a minimum angle above the ground such that if the ball is launched below this angle, it can never clear the bar, no matter how fast it is kicked. What is this angle? (b) If the ball is kicked at \(45.0^{\circ}\) above the horizontal, what must its initial speed be if it is to just clear the bar? Express your answer in \(\mathrm{m} / \mathrm{s}\) and in \(\mathrm{km} / \mathrm{h}\) .
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