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Chapter 3: Problem 72

You passed the salt and pepper shakers to your friend at the other end of a table of height \(0.85 \mathrm{~m}\) by sliding them across the table. They both missed your friend and slid off the table, with velocities of \(5 \mathrm{~m} / \mathrm{s}\) and \(2.5 \mathrm{~m} / \mathrm{s}\), respectively. a) Compare the times it takes the shakers to hit the floor. b) Compare the distance that each shaker travels from the edge of the table to the point it hits the floor.

Short Answer

Expert verified
Answer: The time it takes for both salt and pepper shakers to hit the floor is approximately 0.416 seconds. The salt shaker travels 2.08 meters horizontally, while the pepper shaker travels 1.04 meters horizontally before hitting the floor.

Step by step solution

01

Calculate the time it takes for each shaker to hit the floor

To calculate the time each shaker takes to hit the floor, we can use the vertical motion equation for free-falling objects: \(h_{t} = \frac{1}{2}gt^2\) Where \(h_{t}\) is the vertical distance (table height), \(g\) is acceleration due to gravity (approximately \(9.81 \mathrm{~m/s^2}\)), and \(t\) is the time it takes to hit the floor. Here, we will solve for \(t\) using the given height of the table, which is \(0.85 \mathrm{~m}\): \(0.85 \mathrm{~m} = \frac{1}{2}(9.81 \mathrm{~m/s^2})t^2\) Now, solve for \(t\): \(t = \sqrt{\frac{2\cdot 0.85 \mathrm{~m}}{9.81\mathrm{~m/s^2}}} = \sqrt{0.1731}\) By evaluating the expression, we get: \(t \approx 0.416 \mathrm{~s}\) So, both shakers take approximately \(0.416 \mathrm{~s}\) to hit the floor.
02

Calculate the horizontal distance traveled by each shaker

To calculate the horizontal distance traveled by the shakers, we can use the equation for horizontal motion: \(x_{t} = v_x t\) Here, \(x_{t}\) is the horizontal distance traveled by the shaker, \(v_x\) is the horizontal velocity of the shaker, and \(t\) is the time taken to hit the floor. Since we have already calculated the time to hit the floor (\(t \approx 0.416 \mathrm{~s}\)), we can use the given velocities of the shakers (\(5 \mathrm{~m/s}\) for the salt shaker and \(2.5 \mathrm{~m/s}\) for the pepper shaker) to calculate the horizontal distance traveled by each shaker. For the salt shaker: \(x_{t} = (5 \mathrm{~m/s})(0.416 \mathrm{~s}) = 2.08 \mathrm{~m}\) For the pepper shaker: \(x_{t} = (2.5 \mathrm{~m/s})(0.416 \mathrm{~s})= 1.04 \mathrm{~m}\) a) The times it takes the salt and pepper shakers to hit the floor are the same, approximately \(0.416 \mathrm{~s}\). b) The salt shaker travels \(2.08 \mathrm{~m}\) horizontally from the edge of the table to the point it hits the floor, while the pepper shaker travels \(1.04 \mathrm{~m}\).

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Key Concepts

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

Free-Fall Motion
When an object is subject to gravity alone, it experiences free-fall motion. One crucial aspect of this phenomenon is that the acceleration is constant and equal to the acceleration due to gravity, typically approximated as \( 9.81 \mathrm{~m/s^2} \). During free-fall, horizontal and vertical motions are independent of each other. This means the horizontal velocity of an object has no effect on the vertical motion, and the time it takes for an object to fall is the same for any horizontal velocity as long as it starts from the same height.

In our example, both the salt and pepper shakers fall from a table with a height of \(0.85 \mathrm{~m}\), and despite having different horizontal velocities, they take the same amount of time to hit the ground. It highlights a key characteristic of free-fall motion: all objects, irrespective of their masses and horizontal velocities, fall at the same rate when only gravity acts upon them, neglecting air resistance.
Horizontal Distance Calculation
The crux of calculating the horizontal distance in projectile motion is understanding that horizontal motion is a uniform motion, meaning that there is no acceleration in the horizontal direction (as long as we neglect air resistance). To determine how far an object travels horizontally before landing, we need to know the horizontal velocity and the time the object is in motion.

For the shakers example, we calculate the horizontal distance by multiplying the horizontal velocity \(v_x\) by the time in flight \(t\). Since both shakers share the same flight duration but have different horizontal velocities, we can see that the horizontal distances they cover are directly proportional to their velocities. Thus, the salt shaker, which slips faster, travels a greater horizontal distance than the pepper shaker before they hit the ground.
Kinematic Equations
Kinematic equations allow us to describe the motion characteristics of an object in terms of displacement, velocity, acceleration, and time. These equations are especially helpful for analyzing motion in a straight line under constant acceleration, like in the case of free-fall and horizontal projectile motion.

The equation \( h_{t} = \frac{1}{2}gt^2 \) used in Step 1 of our problem is a kinematic equation that relates the vertical distance fallen (\(h_{t}\)) to the acceleration due to gravity (\(g\)) and the time in motion (\(t\)). Meanwhile, the equation \( x_{t} = v_x t \) used in Step 2 is another kinematic equation that links the horizontal distance (\(x_{t}\)) to the horizontal velocity (\(v_x\)) and the time (\(t\)). By employing these kinematic equations, solving for the unknowns becomes a straightforward task, as long as we interpret each motion—vertical and horizontal—separately.

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Most popular questions from this chapter

A ball is thrown at an angle between \(0^{\circ}\) and \(90^{\circ}\) with respect to the horizontal. Its velocity and acceleration vectors are parallel to each other at a) \(0^{\circ}\) c) \(60^{\circ}\) e) none of the b) \(45^{\circ}\) d) \(90^{\circ}\) above

A baseball is launched from the bat at an angle \(\theta_{0}=30^{\circ}\) with respect to the positive \(x\) -axis and with an initial speed of \(40 \mathrm{~m} / \mathrm{s}\), and it is caught at the same height from which it was hit. Assuming ideal projectile motion (positive \(y\) -axis upward), the velocity of the ball when it is caught is a) \((20.00 \hat{x}+34.64 \hat{y}) \mathrm{m} / \mathrm{s}\). b) \((-20.00 \hat{x}+34.64 \hat{y}) \mathrm{m} / \mathrm{s}\) c) \((34.64 \hat{x}-20.00 \hat{y}) \mathrm{m} / \mathrm{s}\) d) \((34.64 \hat{x}+20.00 \hat{y}) \mathrm{m} / \mathrm{s}\).

A projectile leaves ground level at an angle of \(68^{\circ}\) above the horizontal. As it reaches its maximum height, \(H\), it has traveled a horizontal distance, \(d\), in the same amount of time. What is the ratio \(H / d ?\)

An arrow is shot horizontally with a speed of \(20 \mathrm{~m} / \mathrm{s}\) from the top of a tower \(60 \mathrm{~m}\) high. The time to reach the ground will be a) \(8.9 \mathrm{~s}\) c) \(3.5 \mathrm{~s}\) e) \(1.0 \mathrm{~s}\) b) \(7.1 \mathrm{~s}\) d) \(2.6 \mathrm{~s}\)

To determine the gravitational acceleration at the surface of a newly discovered planet, scientists perform a projectile motion experiment. They launch a small model rocket at an initial speed of \(50.0 \mathrm{~m} / \mathrm{s}\) and an angle of \(30.0^{\circ}\) above the horizontal and measure the (horizontal) range on flat ground to be \(2165 \mathrm{~m}\). Determine the value of \(g\) for the planet.
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