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Chapter 3: Problem 28

(II) An athlete performing a long jump leaves the ground at a 27.0\(^\circ\) angle and lands 7.80 m away. (\(a\)) What was the takeoff speed? (\(b\)) If this speed were increased by just 5.0%, how much longer would the jump be?

Short Answer

Expert verified
(a) The takeoff speed is 9.80 m/s. (b) The jump is 0.72 m longer.

Step by step solution

01

Understand the Problem

We are given the angle of takeoff as \( \theta = 27.0^\circ \), and the horizontal range of the jump as 7.80 m. We need to calculate the initial speed \( v_0 \) for the jump. We will use the jump range formula: \[ R = \frac{v_0^2 \sin(2\theta)}{g} \]where \( R \) is the range, \( \theta \) is the angle of takeoff, and \( g \) is the acceleration due to gravity (\( 9.81 \text{ m/s}^2 \)).
02

Rearrange the Equation to Solve for Initial Velocity

Rearrange the formula to solve for the initial speed \( v_0 \):\[ v_0 = \sqrt{\frac{R \cdot g}{\sin(2\theta)}} \].
03

Calculate the Initial Speed

Substitute the values into the equation:\[ v_0 = \sqrt{\frac{7.80 \cdot 9.81}{\sin(2 \times 27.0^\circ)}} \]Calculate \( \sin(54.0^\circ) \) and find \( v_0 \):\[ v_0 = \sqrt{\frac{76.578}{0.809}} \approx 9.80 \text{ m/s} \].
04

Increase the Initial Speed by 5%

Calculate the new speed after a 5% increase:\[ v_{0,\text{new}} = 9.80 \times 1.05 = 10.29 \text{ m/s} \].
05

Calculate the New Range with Increased Speed

Use the new speed in the range formula:\[ R_{\text{new}} = \frac{10.29^2 \sin(54.0^\circ)}{9.81} \]Calculate the new range:\[ R_{\text{new}} = \frac{105.8841 \times 0.809}{9.81} \approx 8.52 \text{ m} \].
06

Determine the Increase in Jump Distance

Find the difference between the new and the original jump distances:\[ \Delta R = 8.52 - 7.80 = 0.72 \text{ m} \].

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Key Concepts

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

Projectile motion
Projectile motion is a fascinating concept in physics that describes the motion of an object thrown or projected into the air. It is particularly characterized by its two-dimensional motion which is affected by the force of gravity.

When we talk about projectile motion, we consider two main components of motion: horizontal and vertical. These components are independent of each other.
  • The horizontal motion is uniform, meaning it has constant velocity since no acceleration acts along this direction.
  • The vertical motion is accelerated due to gravity, which pulls the object downwards at a rate of 9.81 m/s虏.
Understanding these principles is critical for solving problems related to projectile motion, like figuring out how far or how high a projectile, like a long jump, will go.
Kinematics
Kinematics is the branch of mechanics that focuses on the motion of objects, without considering the forces causing the motion. It's all about the equations of motion.

In projectile motion, kinematics helps us describe and predict an object's path using initial velocity, angles, time in the air, and more. Some common elements studied in kinematics include:
  • Displacement: How far an object moves in a particular direction.
  • Velocity: The speed of an object in a given direction.
  • Acceleration: The rate of change of velocity with time.
Kinematics provides the mathematical tools needed to analyze and solve projectile motion problems, such as finding the initial takeoff speed or the new range when speed changes.
Angle of launch
The angle of launch is a critical factor in projectile motion as it determines both the range and the height of a projectile. The angle is measured from the horizontal plane at which the object is launched.

Different angles lead to different trajectories:
  • A 45掳 angle theoretically provides the maximum range under ideal conditions because it perfectly balances the horizontal and vertical components.
  • Angles less than 45掳 will result in a shorter range but faster flight, while angles greater than 45掳 will heighten the projectile further but shorten its range.
For any specific angle, like the 27掳 in our exercise, the formulation can change the trajectory and thus influence the results, making it vital to understand its impact in kinematic calculations.
Range formula
The range formula is pivotal in calculating how far a projectile will travel horizontally. It relates the range of a projectile with its initial speed, launch angle, and gravitational acceleration.

The formula is given by: \[ R = \frac{v_0^2 \sin(2\theta)}{g} \]Where:
  • \( R \) is the range or horizontal distance travelled.
  • \( v_0 \) is the initial velocity.
  • \( \theta \) is the launch angle.
  • \( g \) is the acceleration due to gravity (9.81 m/s虏).
This formula assumes no air resistance and projects the object from a level ground. In our exercise, the formula helped determine the takeoff speed and predict changes in range due to speed increase, emphasizing its essential role in solving projectile problems.

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Most popular questions from this chapter

A hunter aims directly at a target (on the same level) 38.0 m away. (\(a\)) If the arrow leaves the bow at a speed of 23.1 m/s, by how much will it miss the target? (\(b\)) At what angle should the bow be aimed so the target will be hit?

(II) Extreme-sports enthusiasts have been known to jump off the top of El Capitan, a sheer granite cliff of height 910 m in Yosemite National Park. Assume a jumper runs horizontally off the top of El Capitan with speed 4.0 m/s and enjoys a free fall until she is 150 m above the valley floor, at which time she opens her parachute (Fig. 3-37). (\(a\)) How long is the jumper in free fall? Ignore air resistance. (\(b\)) It is important to be as far away from the cliff as possible before opening the parachute. How far from the cliff is this jumper when she opens her chute?

When Babe Ruth hit a homer over the 8.0-m-high rightfield fence 98 m from home plate, roughly what was the minimum speed of the ball when it left the bat? Assume the ball was hit 1.0 m above the ground and its path initially made a 36\(^\circ\) angle with the ground.

(II) Estimate by what factor a person can jump farther on the Moon as compared to the Earth if the takeoff speed and angle are the same. The acceleration due to gravity on the Moon is one-sixth what it is on Earth.

(III) A diver leaves the end of a 4.0-m-high diving board and strikes the water 1.3 s later, 3.0 m beyond the end of the board. Considering the diver as a particle, determine: (\(a\)) her initial velocity, \(\overrightarrow v_0\) (\(b\)) the maximum height reached; and (\(c\)) the velocity \(\overrightarrow v_f\) with which she enters the water.
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