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Chapter 3: Problem 28

A tiger leaps horizontally from a 7.5 -m-high rock with a speed of \(3.2 \mathrm{~m} / \mathrm{s}\). How far from the base of the rock will she land?

Short Answer

Expert verified
The tiger will land approximately 3.97 meters from the base of the rock.

Step by step solution

01

Identify the Problem Type

This is a projectile motion problem where the tiger leaps horizontally from a certain height with an initial velocity. We need to find the horizontal distance traveled before it lands, which involves using kinematic equations.
02

Determine Time of Flight

The time the tiger spends in the air only depends on the vertical motion. We use the equation for vertical motion: \[ h = \frac{1}{2} g t^2 \]where \( h = 7.5 \, \mathrm{m} \) is the height and \( g = 9.81 \, \mathrm{m/s^2} \) is the acceleration due to gravity. Solve for \( t \) (time):\[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 7.5}{9.81}} \approx 1.24 \, \mathrm{s} \]
03

Calculate Horizontal Distance

The horizontal distance traveled is based on horizontal velocity and time of flight. Use the equation:\[ \text{Horizontal Distance} = v_x \times t \]where \( v_x = 3.2 \, \mathrm{m/s} \). Plug in the values:\[ \text{Horizontal Distance} = 3.2 \, \mathrm{m/s} \times 1.24 \, \mathrm{s} \approx 3.97 \, \mathrm{m} \]
04

Final Answer

The tiger will land approximately \(3.97\, \mathrm{m}\) from the base of the rock.

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Key Concepts

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

Kinematic Equations
Kinematic equations are essential tools in physics that help us describe the motion of objects. These equations connect several key variables: displacement, initial velocity, time, and acceleration.
They are particularly useful in projectile motion problems like the one involving the tiger's leap because they allow us to break down complex motions into solvable steps. In the context of the tiger's leap, the vertical motion is separated from horizontal motion. This separation simplifies calculations and is a common practice in physics.
  • For vertical motion, where the motion is influenced by gravity, we often use the equation: \[ h = \frac{1}{2} g t^2 \] This equation relates the vertical displacement (height), the acceleration due to gravity, and time.
  • For horizontal motion, unaffected by gravity (assuming no air resistance), the horizontal distance is given by another simple equation: \[ \text{Horizontal Distance} = v_x \times t \] Here, \( v_x \) is the constant initial velocity, and \( t \) is time.
Understanding how to apply these equations is key to solving projectile motion problems effectively.
Initial Velocity
Initial velocity is the speed and direction an object has when it begins its motion. In the exercise, the tiger leaps horizontally with an initial velocity of \(3.2 \, \mathrm{m/s}\). This means that, initially, her speed in the horizontal direction is constant because there are no external forces acting in that direction, at least in the ideal physics problem sense.
In projectile motion problems:
  • The horizontal initial velocity remains unchanged through the motion because there鈥檚 no horizontal acceleration (ignoring air resistance).
  • The vertical initial velocity often starts from zero in horizontal leaps, changing only because of the effects of gravity.

This differentiation is crucial in solving problems because it means only the vertical motion is affected by gravity, while the horizontal motion depends entirely on the initial velocity and time.
Acceleration Due to Gravity
Acceleration due to gravity is a fundamental concept in physics. It represents the rate at which an object speeds up as it falls under Earth's gravitational pull. The standard value is approximately \(9.81 \, \mathrm{m/s^2}\). This acceleration affects any object that is in free-fall motion, like the tiger leaping off the rock.
In projectile motion:
  • Gravity only influences the vertical motion, pulling objects downward as they move.
  • It causes the vertical velocity of a falling object to increase at a steady rate.
  • Horizontal motion is generally considered unaffected by gravity, which means it remains constant in ideal conditions without air resistance.

For the tiger's leap, knowing the acceleration allows us to calculate the time it takes to fall from the rock height, making it possible to then determine how far she will travel horizontally.

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Most popular questions from this chapter

Derive a formula for the horizontal range \(R,\) of a projectile when it lands at a height \(h\) above its initial point. (For \(h<0\), it lands a distance \(-h\) below the starting point.) Assume it is projected at an angle \(\theta_{0}\) with initial speed \(v_{0}\).

A light plane is headed due south with a speed relative to still air of \(185 \mathrm{~km} / \mathrm{h}\). After \(1.00 \mathrm{~h}\), the pilot notices that they have covered only \(135 \mathrm{~km}\) and their direction is not south but southeast \(\left(45.0^{\circ}\right) .\) What is the wind velocity?

\((a)\) A long jumper leaves the ground at \(45^{\circ}\) above the horizontal and lands \(8.0 \mathrm{~m}\) away. What is her "takeoff' speed \(v_{0} ?(b)\) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is \(10.0 \mathrm{~m}\) away horizontally and \(2.5 \mathrm{~m},\) vertically below. If she long jumps from the edge of the left bank at \(45^{\circ}\) with the speed calculated in \((a),\) how long, or short, of the opposite bank will she land (Fig. \(3-43\) )?

The specd of a boat in still water is \(v\) . The boat is to make a round trip in a river whose current travels at speed u. Derive a formula for the time needed to make a round trip of total distance \(D\) if the boat makes the round trip by moving (a) upstream and back downstream, and \((b)\) directly across the river and back. We must assume \(u

(II) A motorboat whose speed in still water is 3.40 \(\mathrm{m} / \mathrm{s}\) must aim upstream at an angle of \(19.5^{\circ}\) (with respect to a line perpendicular to the shore) in order to travel dircctly across the stream. (a) What is the spced of the current? (b) What is the resultant speed of the boat with respect to the shore?
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