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Chapter 3: Problem 24

A major-league pitcher can throw a baseball in excess of \(41.0 \mathrm{~m} / \mathrm{s}\). If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is \(17.0 \mathrm{~m}\) away from the point of release?

Short Answer

Expert verified
The ball drops approximately 0.841 meters.

Step by step solution

01

Determine the Time of Flight

To find out how long the baseball is in the air, use the formula for time, given that the ball must travel horizontally 17 meters at a speed of 41 m/s. This can be calculated using the formula: \[ t = \frac{d}{v} \] Substituting the given values: \[ t = \frac{17.0 \, \text{m}}{41.0 \, \text{m/s}} \approx 0.4146 \, \text{s} \] Thus, the time of flight is approximately 0.4146 seconds.
02

Calculate the Vertical Drop

Next, use the time of flight to determine how much the ball falls during this time. Since the ball only falls due to gravity, use the formula for the distance fallen under gravity: \[ y = \frac{1}{2}gt^2 \] where \( g = 9.8 \, \text{m/s}^2 \). Substituting in the values: \[ y = \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (0.4146 \, \text{s})^2 \approx 0.841 \, \text{m} \] So, the ball will drop approximately 0.841 meters.

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Key Concepts

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

Time of Flight
The "Time of Flight" is the duration that an object remains in the air. Imagine throwing a baseball horizontally; from the moment it leaves your hand to when the catcher catches it, that duration is the time of flight. For any projectile, you can determine this using the formula: \[ t = \frac{d}{v} \] Here, \(d\) represents the horizontal distance traveled, and \(v\) is the horizontal velocity.
  • In our example, the ball travels 17 meters with a speed of 41 m/s.
  • Using the formula: \[ t = \frac{17}{41} \approx 0.4146 \, \text{s} \]
This calculation gives us a time of flight of about 0.4146 seconds. Understanding this concept is crucial as it helps us predict how long the ball stays airborne.
Horizontal Velocity
"Horizontal Velocity" refers to the constant speed at which an object moves along a horizontal path. For a baseball thrown by a pitcher, this means the speed across the field without falling yet. The key point is that there is no horizontal acceleration, so the velocity remains consistent.
  • In this scenario, the initial speed is given: 41 m/s.
  • This speed remains constant unless other horizontal forces, like wind, act.
This constant velocity allows us to calculate how long it will take for the ball to cover a specific distance horizontally, which then helps us find the time of flight.
Vertical Drop
The "Vertical Drop" is how much an object falls while it travels horizontally. When a baseball is thrown, gravity pulls it downward even as it moves towards the catcher. To find out how far it drops, we use the following equation: \[ y = \frac{1}{2}gt^2 \] where \(g = 9.8 \, \text{m/s}^2\). This formula tells us how much the ball falls due to gravity during its time of flight.
  • Using our time of flight: 0.4146 seconds
  • Calculate the drop: \[ y = \frac{1}{2} \times 9.8 \times (0.4146)^2 \approx 0.841 \, \text{m} \]
This means the ball will fall approximately 0.841 meters as it travels horizontally. Understanding the vertical drop helps us realize how gravity affects projectiles.
Gravity Effect
"Gravity Effect" is the natural force that causes objects to fall towards the Earth. It acts on all projectiles, pulling them downward. This force is why a baseball thrown horizontally will eventually hit the ground unless caught first. Gravity provides a constant acceleration toward Earth's center: \[ g = 9.8 \, \text{m/s}^2 \] This means that for every second an object is in the air, its downward speed increases by about 9.8 meters per second.
  • The time the ball is airborne (0.4146 seconds) determines how much speed it gains downward.
  • This acceleration causes the vertical drop from the "Vertical Drop" section.
Recognizing the impact of gravity helps explain why even horizontally thrown objects don't move in a straight line but instead follow a curved path.

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Most popular questions from this chapter

A small can is hanging from the ceiling. A rifle is aimed directly at the can, as the figure illustrates. At the instant the gun is fired, the can is released. Ignore air resistance and show that the bullet will always strike the can, regardless of the initial speed of the bullet. Assume that the bullet strikes the can before the can reaches the ground.

As preparation for this problem, review Conceptual Example \(10 .\) The drawing shows two planes each about to drop an empty fuel tank. At the moment of release each plane has the same speed of \(135 \mathrm{~m} / \mathrm{s}\), and each tank is at the Plane \(\mathrm{B}\) same height of \(2.00 \mathrm{~km}\) above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of \(15.0^{\circ}\) above the horizontal and the other is flying at an angle of \(15.0^{\circ}\) below the horizontal. Find the magnitude and direction of the velocity with which the fuel tank hits the ground if it is from (a) plane A and (b) plane B. In each part, give the directional angles with respect to the horizontal.

A golf ball rolls off a horizontal cliff with an initial speed of \(11.4 \mathrm{~m} / \mathrm{s} .\) The ball falls a vertical distance of \(15.5 \mathrm{~m}\) into a lake below. (a) How much time does the ball spend in the air? (b) What is the speed \(v\) of the ball just before it strikes the water?

Concept Simulation 3.2 at reviews the concepts that are important in this problem. A golfer imparts a speed of \(30.3 \mathrm{~m} / \mathrm{s}\) to a ball, and it travels the maximum possible distance before landing on the green. The tee and the green are at the same elevation. (a) How much time does the ball spend in the air? (b) What is the longest "hole in one" that the golfer can make, if the ball does not roll when it hits the green?

In the javelin throw at a track-and-field event, the javelin is launched at a speed of \(29 \mathrm{~m} /\) s at an angle of \(36^{\circ}\) above the horizontal. As the javelin travels upward, its velocity points above the horizontal at an angle that decreases as time passes. How much time is required for the angle to be reduced from \(36^{\circ}\) at launch to \(18^{\circ}\) ?
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