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Playing shortstop, you pick up a ground ball and throw it to second base. The ball is thrown horizontally, with a speed of \(22 \mathrm{~m} / \mathrm{s}\), directly toward point \(A\) as shown in Figure 4.37. When the ball reaches the second baseman \(0.45 \mathrm{~s}\) later, it is caught at point B. (a) How far were you from the second baseman? (b) What is the distance of vertical drop, from \(A\) to \(B\) ?

Short Answer

Expert verified
(a) 9.9 m; (b) 0.99 m drop.

Step by step solution

01

Identify Given Information

We know the following information:- Horizontal velocity of the ball, \(v = 22 \text{ m/s}\)- Time for the ball to reach the second baseman, \(t = 0.45 \text{ s}\)
02

Calculate Horizontal Distance

The horizontal distance \(d\) can be calculated using the formula for constant speed: \[d = v \times t\]Substitute the known values:\(d = 22 \text{ m/s} \times 0.45 \text{ s} = 9.9 \text{ m}\)So, you were \(9.9\) meters away from the second baseman.
03

Determine Vertical Drop

To find the vertical drop, we need to use the formula for the distance fallen in free-fall, \(h = \frac{1}{2} g t^2\), where \(g\) is the acceleration due to gravity \(9.8 \text{ m/s}^2\).\[h = \frac{1}{2} \times 9.8 \text{ m/s}^2 \times (0.45 \text{ s})^2\]\(h \approx \frac{1}{2} \times 9.8 \times 0.2025 = 0.99 \text{ m}\)So, the ball vertically dropped approximately \(0.99\) meters from \(A\) to \(B\).

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Key Concepts

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

Horizontal Velocity
Horizontal velocity refers to the constant speed at which an object moves along a horizontal path. It is a fundamental factor in projectile motion, where an object is launched into the air and follows a curved trajectory.
In the case of the baseball, the horizontal velocity is given as \(22 \text{ m/s}\), meaning that the ball travels at a constant speed of \(22 \text{ m/s}\) in the horizontal direction. This speed does not change as it moves towards the second baseman.
To find out how far the baseball travels, we use the formula for constant speed:
  • Distance \(d = v \times t\)
With \(v\) being the horizontal velocity and \(t\) the time taken. Substituting \(22 \text{ m/s}\) for the speed and \(0.45 \text{ s}\) for the time, the ball travels a distance of \(9.9 \text{ meters}\) horizontally. This calculation shows how horizontal velocity and time work together to determine distance.
Vertical Drop
The vertical drop in projectile motion refers to how much an object falls vertically due to gravity, while it moves horizontally. This drop is caused by the force of gravity acting on the object during its flight.
Even though the baseball is thrown horizontally, it simultaneously falls towards the ground. We calculate the vertical drop using the formula for free fall:
  • Vertical drop \(h = \frac{1}{2} g t^2\)
Here, \(g\) represents the acceleration due to gravity, which is \(9.8 \text{ m/s}^2\). By substituting \(0.45 \text{ s}\) for the time, the equation yields\[ h = \frac{1}{2} \times 9.8 \times (0.45)^2 \approx 0.99 \text{ m} \]This tells us that while the baseball travels horizontally, it drops approximately \(0.99 \text{ meters}\) vertically, demonstrating the simultaneous horizontal and vertical components of projectile motion.
Acceleration Due to Gravity
Acceleration due to gravity, often denoted by \(g\), is a key component in understanding projectile motion. It represents the rate at which an object accelerates towards the Earth's surface due to gravitational force.
On Earth, this value is roughly \(9.8 \text{ m/s}^2\). This means that in the absence of other forces, an object's velocity increases by \(9.8 \text{ m/s}\) every second it is in free fall.

Impact on the Baseball

In the exercise, gravity is what causes the baseball to drop \(0.99 \text{ meters}\) as it moves horizontally. Although the ball is thrown horizontally, it cannot overcome the pull of gravity, resulting in a vertical descent.
Gravity acts independently of any horizontal forces, causing the vertical component of motion to appear separate from the horizontal motion. Therefore, while the horizontal velocity remains constant, the vertical drop is dictated solely by gravity's pull on the ball, highlighting the dual nature of projectile trajectories.

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Most popular questions from this chapter

A crow is flying horizontally with a constant speed of \(2.70 \mathrm{~m} / \mathrm{s}\) when it releases a clam from its beak as shown in Figure 4.38. The clam lands on the rocky beach \(2.10 \mathrm{~s}\) later. Just before the clam lands, what is (a) its horizontal component of velocity and (b) its vertical component of velocity?

A passenger walks from one side of a ferry to the other as it approaches a dock. If the passenger's velocity is \(1.50 \mathrm{~m} / \mathrm{s}\) due north relative to the ferry and \(4.50 \mathrm{~m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) west of north relative to the water, what are the direction and magnitude of the ferry's velocity relative to the water?

As an airplane taxis on the runway with a speed of \(16.5 \mathrm{~m} / \mathrm{s}\), a flight attendant walks toward the tail of the plane with a speed of \(1.22 \mathrm{~m} / \mathrm{s}\). What is the flight attendant's speed relative to the ground?

Vector \(\overrightarrow{\mathbf{A}}\) has a magnitude of \(50 \mathrm{~km}\) and points in the positive \(x\) direction. A second vector, \(\overrightarrow{\mathbf{B}}\), has a magnitude of \(120 \mathrm{~km}\) and points at an angle of \(70^{\circ}\) below the \(x\) axis. Which vector has (a) the greater \(x\) component and (b) the greater \(y\) component?

A softball is thrown from the origin of an \(x-y\) coordinate system with an initial speed of \(18 \mathrm{~m} / \mathrm{s}\) at an angle of \(35^{\circ}\) above the horizontal. (a) Find the \(x\) and \(y\) positions of the softball at the times \(t=0.50 \mathrm{~s}, 1.0 \mathrm{~s}, 1.5 \mathrm{~s}\), and \(2.0 \mathrm{~s}\). (b) Plot the results from part (a) on an \(x-y\) coordinate system, and sketch the parabolic curve that passes through them.

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