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Chapter 4: Problem 55

A ball rolls horizontally off the top of a stairway with a speed of \(1.52 \mathrm{~m} / \mathrm{s}\). The steps are \(20.3 \mathrm{~cm}\) high and \(20.3 \mathrm{~cm}\) wide. Which step does the ball hit first?

Short Answer

Expert verified
The ball hits the second step first.

Step by step solution

01

Converting Units

First, convert the step dimensions from centimeters to meters to be consistent with the other units. Since there are 100 centimeters in a meter, each step is \(0.203\) meters high and \(0.203\) meters wide.
02

Obtaining Time of Flight

Since the ball rolls off the stairs horizontally, its vertical motion can be analyzed independently of horizontal motion. Use the equation for vertical motion: \( y = \frac{1}{2}gt^2 \), where \( y = 0.203 \) meters is the height of a step, and \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity. Solve for \( t \): \( t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2 \cdot 0.203}{9.81}} \approx 0.203\, \text{s}.\) This is the time taken to fall one height.
03

Calculating Horizontal Distance

The horizontal distance covered in time \( t \) is given by \( x = vt \), where \( v = 1.52 \, \text{m/s} \) is the initial horizontal speed of the ball. Therefore, \( x = 1.52 \times 0.203 \approx 0.309 \) meters. This is how far the ball travels horizontally during the fall time for one step.
04

Determining the Step Hit First

To find which step the ball hits, compare the horizontal distance \( x \) to the width of a step. Since each step is \(0.203\) meters wide, divide the horizontal distance by the step width: \( n = \frac{x}{\text{width}} = \frac{0.309}{0.203} \approx 1.52. \) This means the ball first hits between the second step and the start of the third step.

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Key Concepts

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

Horizontal Motion
In projectile motion, horizontal motion is an essential aspect where an object moves parallel to the ground. When a ball rolls off a stairway, it carries with it an initial horizontal speed, which is unchanging, as there are no horizontal forces acting (if we ignore air resistance). This constant speed competes with gravity pulling the ball downwards. For the problem at hand, the ball begins its motion with a speed of \(1.52 \, \mathrm{m/s}\). This consistent rate is applied through its entire flight until the ball meets an obstacle like the ground or a step.
  • Horizontal velocity remains constant throughout the motion.
  • No acceleration affects horizontal motion directly.
  • Measured by multiplying horizontal speed by the time of flight.
Understanding horizontal motion helps predict where the ball lands horizontally, based on the time it spends airborne.
Vertical Motion
Vertical motion differs from horizontal motion as it is constantly influenced by gravity, resulting in acceleration. For a ball moving downward off a platform like stairs, gravity accelerates it at a rate of \(9.81 \, \mathrm{m/s^2}\). The vertical motion is independent of the horizontal, meaning they do not affect each other.

Key points:
  • Vertical motion follows a free-fall path.
  • Velocity increases linearly due to constant acceleration.
  • Formulated as \(y = \frac{1}{2}gt^2\), where \(y\) is the vertical distance.
By using the vertical motion equation, one can solve for the time it takes for the ball to hit a certain distance vertically, which helps in determining its time of flight.
Acceleration Due to Gravity
Gravity is crucial in influencing vertical motion, imparting an unrelenting pull towards the Earth. This pull is quantified by acceleration due to gravity, denoted as \( g \) which is approximately \(9.81 \, \mathrm{m/s^2}\). It acts equally on all objects, non-discriminant of their mass, in the absence of air resistance.
  • Gravity acts on vertical motion only.
  • Causes objects to accelerate as they descend.
  • Determines the rate at which the speed increases vertically.
By utilizing \( g \) in equations like \(y = \frac{1}{2} gt^2\), you can determine how long it takes for an object to fall a certain vertical distance. This makes gravity a key factor in predicting projectile motion outcomes.
Time of Flight
The time of flight refers to the total time an object spends in the air from the moment it begins its motion till it makes contact with the ground or another object. For the ball rolling off the stairs, this time is calculated by assessing the vertical motion. By rearranging and solving the formula \(y = \frac{1}{2} gt^2\), we determine the time it takes for the ball to fall a height equal to one step.
  • Calculated based on vertical distance and gravity.
  • Important for predicting where the object lands horizontally.
  • Requires eliminating horizontal influences in the calculation.
The calculated \(0.203 \, \mathrm{s}\), accounts for the ball's drop from the stair's top which helps in finding how far it travels horizontally before hitting a step.

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Most popular questions from this chapter

A football player punts the football so that it will have a "hang time" (time of flight) of \(4.5 \mathrm{~s}\) and land \(46 \mathrm{~m}\) away. If the ball leaves the player's foot \(150 \mathrm{~cm}\) above the ground, what must be the (a) magnitude and (b) angle (relative to the horizontal) of the ball's initial velocity?

A rifle that shoots bullets at \(460 \mathrm{~m} / \mathrm{s}\) is to be aimed at a target \(45.7 \mathrm{~m}\) away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that the bullet hits dead center?

At \(t_{1}=2.00 \mathrm{~s}\), the acceleration of a particle in counterclockwise circular motion is \(\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(4.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). It moves at constant speed. At time \(t_{2}=5.00 \mathrm{~s}\), the particle's acceleration is \(\left(4.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(-6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). What is the radius of the path taken by the particle if \(t_{2}-t_{1}\) is less than one period?

What is the magnitude of the acceleration of a sprinter running at \(10 \mathrm{~m} / \mathrm{s}\) when rounding a turn of radius \(25 \mathrm{~m}\) ?

An astronaut is rotated in a horizontal centrifuge at a radius of \(5.0 \mathrm{~m}\). (a) What is the astronaut's speed if the centripetal acceleration has a magnitude of \(7.0 \mathrm{~g} ?\) (b) How many revolutions per minute are required to produce this acceleration? (c) What is the period of the motion?
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