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Chapter 3: Problem 28

Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.9 m/s parallel to the ground. Upon contact with the bat the ball is 1.2 m above the ground. Player B wishes to duplicate this bunt, in so far as he also wants to give the ball a velocity parallel to the ground and have his ball travel the same horizontal distance as player A鈥檚 ball does. However, player B hits the ball when it is 1.5 m above the ground. What is the magnitude of the initial velocity that player B鈥檚 ball must be given?

Short Answer

Expert verified
Player B needs an initial velocity of 1.7 m/s.

Step by step solution

01

Determine the Time of Flight for Player A

To find the time it takes for the ball hit by Player A to reach the ground, use the vertical motion equation:\[ s = ut + \frac{1}{2}gt^2 \]where \( s \) is the vertical distance (1.2 m), \( u \) is the initial vertical velocity (0 m/s), and \( g \) is the acceleration due to gravity (9.8 m/s虏). Solving for \( t \) gives:\[ 1.2 = 0 + \frac{1}{2}(9.8)t^2 \]\[ t^2 = \frac{2 \times 1.2}{9.8} \]\[ t \approx 0.495 \text{ s} \]
02

Calculate the Horizontal Distance for Player A

Using the horizontal motion formula, since the velocity is constant and horizontal:\[ d = v \cdot t \]where \( d \) is the horizontal distance, \( v \) is the horizontal velocity (1.9 m/s), and \( t \) is the time (0.495 s), we have:\[ d = 1.9 \cdot 0.495 \]\[ d \approx 0.9405 \text{ m} \]
03

Determine the Time of Flight for Player B

Player B hits the ball from a height of 1.5 m above the ground. Use the same vertical equation:\[ s = ut + \frac{1}{2}gt^2 \]where \( s = 1.5 \) m. Solving for \( t \):\[ 1.5 = 0 + \frac{1}{2}(9.8)t^2 \]\[ t^2 = \frac{2 \times 1.5}{9.8} \]\[ t \approx 0.553 \text{ s} \]
04

Calculate Required Horizontal Velocity for Player B

To achieve the same horizontal distance as Player A, Player B's ball must travel 0.9405 m in 0.553 s. Using the formula:\[ d = v \cdot t \]Re-arrange to solve for the horizontal velocity \( v \):\[ v = \frac{d}{t} \]\[ v = \frac{0.9405}{0.553} \]\[ v \approx 1.7 \text{ m/s} \]
05

Conclusion

Therefore, the initial horizontal velocity required for Player B to duplicate the bunt's horizontal distance is approximately 1.7 m/s.

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Key Concepts

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

Time of Flight
In projectile motion, the **Time of Flight** refers to the duration a projectile is in the air after launch. It depends on two main factors:
  • Initial height: The higher the ball is launched from, the longer it will take to hit the ground.
  • Gravity: Gravity pulls the ball downward, affecting how long it stays in the air.
To find the time of flight for Player A and Player B's baseball, we use the equation:\[ s = rac{1}{2}gt^2 \]where:
  • \( s \) = the vertical distance (initial height), and
  • \( g \) = acceleration due to gravity.
This means no initial vertical velocity because the ball is hit parallel to the ground.By solving this equation, you can find how long it takes for the ball to fall to the ground.
Horizontal Distance
The **Horizontal Distance** in projectile motion refers to the total distance the projectile travels along the horizontal axis during its flight. This metric is crucial for players like Player A and Player B who aim to duplicate a specific hitting style in baseball.
To calculate this distance, we utilize the formula:\[ d = v \, \cdot \, t \]where:
  • \( d \) = horizontal distance,
  • \( v \) = constant horizontal velocity, and
  • \( t \) = time of flight.
The horizontal velocity stays the same because no forces are acting horizontally (ignoring air resistance for simplicity). The longer the time of flight, the greater the horizontal distance traveled, assuming constant velocity.
Initial Velocity
The **Initial Velocity** in this context refers to the speed given to the baseball at the moment it is struck. For Player A, the initial velocity is provided as 1.9 m/s, purely horizontal. The challenge for Player B is to find a similar initial velocity that allows the baseball to reach the same horizontal distance even from a different height.
  • Player B aims to achieve the same distance by adjusting the initial velocity due to the difference in height from when the ball is hit.
  • Calculating this requires rearranging the horizontal distance formula to solve for initial velocity.
In this case, the initial velocity for Player B must be about 1.7 m/s to match Player A's horizontal performance.
Gravity
**Gravity** is a fundamental force in physics that impacts all projectile motions. On Earth, gravity pulls objects downward towards the ground at an acceleration rate of approximately 9.8 m/s虏. In the context of our exercise:
  • Gravity influences the time it takes for a baseball to fall to the ground. Thus it directly affects the time of flight.
  • It does not influence the horizontal distance covered but affects the vertical motion of the ball.
Understanding gravity's role allows us to calculate how long objects will stay airborne and enables the precise prediction of landing points.

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Most popular questions from this chapter

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