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Chapter 5: Problem 2

A Small Ball A small ball rolls horizontally off the edge of a tabletop that is \(1.20 \mathrm{~m}\) high. It strikes the floor at a point \(1.52 \mathrm{~m}\) horizontally away from the edge of the table. (a) How long is the ball in the air? (b) What is its speed at the instant it leaves the table?

Short Answer

Expert verified
The ball is in the air for approximately 0.495 s and its speed when it leaves the table is approximately 3.07 m/s.

Step by step solution

01

- Determine the time the ball is in the air (Vertical Motion)

Use the kinematic equation for vertical motion to determine the time it takes for the ball to fall to the floor. The equation is \[ y = v_{y0} t + \frac{1}{2} g t^2 \] where: - \( y = 1.20 \text{ m} \) (height of the table) - Initial vertical velocity \( v_{y0} = 0 \text{ m/s} \) - \( g = 9.8 \text{ m/s}^2 \) (acceleration due to gravity) Plugging in the values, we get: \[ 1.20 = 0 \cdot t + \frac{1}{2} (9.8) t^2 \] Solving for \( t \), we get: \[ t^2 = \frac{2 \times 1.20}{9.8} \] \[ t^2 = \frac{2.4}{9.8} \] \[ t^2 = 0.2449 \] \[ t = \sqrt{0.2449} \] \[ t \approx 0.495 \text{ s} \] So, the time the ball is in the air is approximately \( 0.495 \text{ s} \).
02

- Determine the horizontal speed of the ball

Use the horizontal distance to find the horizontal speed. The formula is \[ x = v_{x} t \] where: - \( x = 1.52 \text{ m} \) (horizontal distance) - \( t = 0.495 \text{ s} \) Plugging in the values, we get: \[ 1.52 = v_{x} \times 0.495 \] Solving for \( v_{x} \), we get: \[ v_{x} = \frac{1.52}{0.495} \] \[ v_{x} \approx 3.07 \text{ m/s} \] So, the speed of the ball at the instant it leaves the table is approximately \( 3.07 \text{ m/s} \).

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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 used to describe the motion of an object. These equations relate the displacement, initial velocity, final velocity, acceleration, and time. There are four primary kinematic equations for uniformly accelerated motion. In the context of projectile motion, we typically use the following kinematic equation for vertical motion:





In our exercise, we use this equation to find out how long the ball is in the air. By focusing on vertical motion, we know the initial vertical velocity is zero. Hence,
Horizontal Motion
Horizontal motion in projectile motion is straightforward because there is no horizontal acceleration. The horizontal velocity remains constant throughout the motion. The formula to describe horizontal displacement is:

This tells us we can easily determine the horizontal speed of the ball by dividing the total horizontal distance by the time it spends in the air. This is helpful as it simplifies one part of the problem considerably.
Vertical Motion
Vertical motion is influenced by gravity, which causes acceleration downwards. In our exercise, the ball starts with zero initial vertical velocity since it rolls horizontally off the table. Therefore, the vertical motion equation simplifies to:

Solving for Time







We substituted the vertical height and solved for time, which gave us approximately 0.495 seconds.
Acceleration Due to Gravity
Gravity causes a uniformly accelerated motion downwards with an acceleration of approximately 9.8 m/s虏 on Earth's surface. In projectile motion, this constant force only affects vertical motion, pulling objects downward.

In the context of our problem, gravity helps us determine the time the ball is in the air by using the kinematic equation for vertical motion. This value of 9.8 m/s虏 is crucial for finding out how long the ball takes to hit the floor.

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Most popular questions from this chapter

. Stairway 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?

An Earth Satellite An Earth satellite moves in a circular orbit \(640 \mathrm{~km}\) above Earth's surface with a period of \(98.0 \mathrm{~min}\). What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?

Soccer Ball A soccer ball is Problem 17 . kicked from the ground with an initial speed of \(19.5 \mathrm{~m} / \mathrm{s}\) at an upward angle of \(45^{\circ} .\) A player \(55 \mathrm{~m}\) away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground? Neglect air resistance. 19\. Stairway 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?

(a) What is the magnitude of the centripetal acceleration of an object on Earth's equator due to the rotation of Earth? (b) What would the period of rotation of Earth have to be for objects on the equator to have a centripetal acceleration with a magnitude of \(9.8 \mathrm{~m} / \mathrm{s}^{2} ?\)

Dart A dart is thrown horizontally with an initial speed of \(10 \mathrm{~m} / \mathrm{s}\) toward point \(P\), the bull's-eye on a dart board. It hits at point \(Q\) on the rim, vertically below \(P 0.19\) s later. (a) What is the distance \(P Q ?\) (b) How far away from the dart board is the dart released?
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