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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

Identify Does the magnitude of a vector refer to its length or its direction?

Tower An air traffic controller observes two airplanes approaching the airport. The displacement from the control tower to plane 1 is given by the vector \(\overrightarrow{\mathbf{A}}\), which has a magnitude of \(220 \mathrm{~km}\) and points in a direction \(32^{\circ}\) north of west. The displacement from the control tower to plane 2 is given by the vector \(\overrightarrow{\mathbf{B}}\), which has a magnitude of \(140 \mathrm{~km}\) and points \(65^{\circ}\) east of north. (a) Sketch the vectors \(\overrightarrow{\mathbf{A}},-\overrightarrow{\mathbf{B}}\), and \(\overrightarrow{\mathbf{D}}=\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\). Notice that \(\vec{D}\) is the displacement from plane 2 to plane 1 . (b) Use components to find the magnitude and the direction of the vector \(\overrightarrow{\mathbf{D}}\).

Which of the following shapes is seen in the path of a projectile: straight line, parabola, circle, ellipse, hyperbola?

What distinguishes a vector from a scalar?

Vector \(\overline{\mathbf{A}}\) points in the negative \(y\) direction and has a magnitude of \(5 \mathrm{~km}\). Vector \(\overrightarrow{\mathbf{B}}\) has a magnitude of \(15 \mathrm{~km}\) and points in the positive \(x\) direction. Use components to find the magnitude of (a) \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\), (b) \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\), and (c) \(\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}\).

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