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Chapter 4: Problem 22

A small ball rolls horizontally off the edge of a tabletop that is \(1.50 \mathrm{~m}\) high. It strikes the floor at a point \(1.52 \mathrm{~m}\) horizontally from the table edge. (a) How long is the ball in the air? (b) What is its sneed at the instant it leaves the table?

Short Answer

Expert verified
(a) The ball is in the air for about 0.553 seconds. (b) The speed when it leaves the table is approximately 2.75 m/s.

Step by step solution

01

Analyze the Problem

The ball falls off a table and hits the ground. We can consider the horizontal and vertical motions separately, using equations of motion. The vertical motion is a free-fall, and the horizontal motion has constant speed. We will first find the time of flight using the vertical motion.
02

Solve for Time of Flight

For vertical motion, the ball falls from a height of 1.50 m. Using the equation of motion for free fall, we have: \( y = \frac{1}{2} g t^2 \), where \( y = 1.50 \) m and \( g = 9.81 \) m/s").Solving for \( t \):\[ 1.50 = \frac{1}{2} \times 9.81 \times t^2 \]\[ t^2 = \frac{1.50 \times 2}{9.81} \]\[ t = \sqrt{\frac{3.00}{9.81}} \approx 0.553 \text{ seconds} \]
03

Determine Horizontal Velocity

Now, use the horizontal distance to find the initial speed. The horizontal motion equation is \( x = v_x t \), where \( x = 1.52 \) m.So, \( v_x = \frac{x}{t} = \frac{1.52}{0.553} \approx 2.75 \text{ m/s} \). This is the speed of the ball when it leaves the table.

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Key Concepts

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

Free Fall
When an object experiences free fall, it moves under the influence of gravity alone. This means no other forces, like air resistance, act on the object. An important characteristic of free fall is that all objects fall at the same rate irrespective of their masses. The acceleration due to gravity (
  • Symbol: usually denoted as \( g \)
  • Standard value: \( g = 9.81 \ \text{m/s}^2 \)
In the context of our exercise, the small ball falls freely from a height of 1.50 meters. When analyzing its free fall, we used the equation of motion: \[ y = \frac{1}{2} g t^2 \]Here, \( y \) represents the vertical distance (1.50 meters), while \( t \) is the time the ball is in the air. This equation helps us understand how the ball accelerates downward, leading to the solution that it is in the air for approximately 0.553 seconds.
Horizontal Motion
In projectile motion, horizontal motion is considered separately from vertical motion. It occurs at a constant speed because, typically, there's no horizontal acceleration. The main formula to understand horizontal motion is:\[ x = v_x t \]
  • \( x \) is the horizontal distance traveled, which in our problem is 1.52 meters.
  • \( v_x \) is the horizontal speed.
For the ball in the exercise, since there's no horizontal force acting on it, its speed remains unchanged. This means the horizontal speed by which the ball leaves the tabletop is the same as when it lands. By using the calculated time of flight \( t = 0.553 \ \text{s} \), you find that \( v_x \) is approximately 2.75 m/s.
Equations of Motion
Equations of motion are mathematical equations that describe the movement of objects. These equations are crucial as they allow us to predict the future position and velocity of moving objects. They are split into vertical and horizontal components for projectile motion problems, like in our exercise.The main equations involved are:
  • Vertical: \[ y = \frac{1}{2} g t^2 \]This equation determines how far an object falls under the force of gravity.
  • Horizontal:\[ x = v_x t \]This equation calculates the horizontal distance covered by an object moving at constant speed.
These equations allowed us to solve for the unknowns in our problem: calculating how long the ball stayed in the air and determining the speed it had when it left the table. Understanding each equation's application will aid in solving similar problems.
Time of Flight
The time of flight is a crucial aspect of projectile motion which tells how long an object stays in motion before it hits the ground. It's determined by the vertical motion of the object, impacted by gravity.For the ball in the exercise, we calculated the time of flight using the equation:\[ y = \frac{1}{2} g t^2 \]Here, solving for \( t \) provided us with a time of 0.553 seconds. This calculation shows how gravity influences the duration the ball remains in the air, independent of its horizontal speed.Learning to calculate the time of flight can help predict when and where an object will land, which is essential in various real-world applications, from sports to engineering.

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Most popular questions from this chapter

Ship \(A\) is located \(4.0 \mathrm{~km}\) north and \(2.5 \mathrm{~km}\) east of ship \(B\). Ship \(A\) has a velocity of \(22 \mathrm{~km} / \mathrm{h}\) toward the south, and \(\operatorname{ship} B\) has a velocity of \(40 \mathrm{~km} / \mathrm{h}\) in a direction \(37^{\circ}\) north of east. (a) What is the velocity of \(A\) relative to \(B\) in unit-vector notation with \(\hat{i}\) toward the east? (b) Write an expression (in terms of \(\hat{\mathrm{i}}\) and \(\hat{\mathrm{j}}\) ) for the position of \(A\) relative to \(B\) as a function of \(t\), where \(t=0\) when the ships are in the positions described above. (c) At what time is the separation between the ships least? (d) What is that least separation?

From the origin, a particle starts at \(t=0 \mathrm{~s}\) with a velocity \(\vec{v}=7.0 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}\) and moves in the \(x y\) plane with a constant acceleration of \(\vec{a}=(-9.0 \hat{\mathrm{i}}+3.0 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}^{2} .\) At the time the particle reaches the maximum \(x\) coordinate, what is its (a) velocity and (b) position vector?

A projectile's launch speed is \(6.00\) times that of its speed at its maximum height. Find the launch angle \(\theta_{0}\).

During a tennis match, a player serves the ball at \(23.6 \mathrm{~m} / \mathrm{s}\), with the center of the ball leaving the racquet horizontally \(2.42 \mathrm{~m}\) above the court surface. The net is \(12 \mathrm{~m}\) away and \(0.90 \mathrm{~m}\) high. When the ball reaches the net, (a) does the ball clear it and (b) what is the distance between the center of the ball and the top of the net? Suppose that, instead, the ball is served as before but now it leaves the racquet at \(5.00^{\circ}\) below the horizontal. When the ball reaches the net, (c) does the ball clear it and (d) what now is the distance between the center of the ball and the top of the net?

A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wall to break apart the wall. The machine was not placed near the wall because then arrows could reach it from the castle wall. Instead, it was positioned so that the stone hit the wall during the second half of its flight. Suppose a stone is launched with a speed of \(v_{0}=30.0\) \(\mathrm{m} / \mathrm{s}\) and at an angle of \(\theta_{0}=40.0^{\circ}\). What is the speed of the stone if it hits the wall (a) just as it reaches the top of its parabolic path and (b) when it has descended to half that height? (c) As a percentage, how much faster is it moving in part (b) than in part (a)?
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