A Student Fires A Cannonball Horizontally With A Speed Of $20.0 , \text{m/s}$ From A Height Of $68.0 , \text{m}$. Neglect Air Resistance.1. How Long Did The Ball Remain In The Air? - $\square , \text{units}$2. How

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Introduction

In physics, understanding the motion of objects under the influence of gravity is crucial. When a cannonball is fired horizontally from a certain height, it follows a parabolic trajectory under the sole influence of gravity. In this article, we will explore the motion of a cannonball fired horizontally with a speed of $20.0 , \text{m/s}$ from a height of $68.0 , \text{m}$. We will neglect air resistance and focus on calculating the time the ball remains in the air.

The Horizontal and Vertical Components of Motion

When the cannonball is fired horizontally, it has an initial horizontal velocity of $20.0 , \text{m/s}$. Since there are no forces acting horizontally, the horizontal velocity remains constant throughout the motion. The vertical component of the motion, however, is influenced by gravity. The acceleration due to gravity is $9.8 , \text{m/s}^2$, acting in the downward direction.

Time of Flight

To calculate the time the ball remains in the air, we need to consider the vertical component of the motion. The initial vertical velocity is zero, and the acceleration due to gravity is $9.8 , \text{m/s}^2$. We can use the equation of motion under constant acceleration to find the time of flight:

s=ut+12at2s = ut + \frac{1}{2}at^2

where $s$ is the displacement, $u$ is the initial velocity, $t$ is the time, and $a$ is the acceleration.

In this case, the initial vertical velocity is zero, and the displacement is the height from which the ball is fired, which is $68.0 , \text{m}$. The acceleration due to gravity is $9.8 , \text{m/s}^2$. We can rearrange the equation to solve for time:

t=2sat = \sqrt{\frac{2s}{a}}

Substituting the values, we get:

t=2×68.0 m9.8 m/s2t = \sqrt{\frac{2 \times 68.0 \, \text{m}}{9.8 \, \text{m/s}^2}}

t=136.0 m9.8 m/s2t = \sqrt{\frac{136.0 \, \text{m}}{9.8 \, \text{m/s}^2}}

t=13.87 s2t = \sqrt{13.87 \, \text{s}^2}

t=3.72 st = 3.72 \, \text{s}

Conclusion

In conclusion, the cannonball fired horizontally with a speed of $20.0 , \text{m/s}$ from a height of $68.0 , \text{m}$ will remain in the air for approximately $3.72 , \text{s}$. This calculation assumes that air resistance is negligible and that the only force acting on the ball is gravity.

Discussion

The time of flight of the cannonball is determined by the vertical component of its motion, which is influenced by gravity. The horizontal component of the motion remains constant throughout the flight. This problem illustrates the importance of considering both the horizontal and vertical components of motion when analyzing the trajectory of an object under the influence of gravity.

Key Points

  • The time of flight of the cannonball is determined by the vertical component of its motion.
  • The horizontal component of the motion remains constant throughout the flight.
  • The acceleration due to gravity is $9.8 , \text{m/s}^2$, acting in the downward direction.
  • The initial vertical velocity is zero, and the displacement is the height from which the ball is fired.

Further Reading

For further reading on the topic of motion under gravity, we recommend the following resources:

  • [1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics. John Wiley & Sons.
  • [2] Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers. Cengage Learning.

References

[1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics. John Wiley & Sons.

[2] Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers. Cengage Learning.

Introduction

In our previous article, we explored the motion of a cannonball fired horizontally with a speed of $20.0 , \text{m/s}$ from a height of $68.0 , \text{m}$. We calculated the time the ball remains in the air, assuming that air resistance is negligible and that the only force acting on the ball is gravity. In this article, we will answer some frequently asked questions related to this problem.

Q&A

Q1: What is the significance of the horizontal component of motion in this problem?

A1: The horizontal component of motion remains constant throughout the flight, which means that the cannonball will travel a horizontal distance equal to its initial horizontal velocity multiplied by the time of flight.

Q2: How does the acceleration due to gravity affect the time of flight?

A2: The acceleration due to gravity is $9.8 , \text{m/s}^2$, acting in the downward direction. This acceleration causes the cannonball to accelerate downward, which in turn affects the time of flight.

Q3: What is the relationship between the time of flight and the height from which the cannonball is fired?

A3: The time of flight is directly proportional to the square root of the height from which the cannonball is fired. This means that if the height is increased, the time of flight will also increase.

Q4: How does air resistance affect the motion of the cannonball?

A4: Air resistance is neglected in this problem, which means that the only force acting on the ball is gravity. In reality, air resistance would slow down the cannonball and affect its trajectory.

Q5: What is the significance of the initial vertical velocity in this problem?

A5: The initial vertical velocity is zero, which means that the cannonball is fired horizontally. This is why the time of flight is determined solely by the vertical component of motion.

Q6: How can the time of flight be calculated if the initial vertical velocity is not zero?

A6: If the initial vertical velocity is not zero, the time of flight can be calculated using the equation of motion under constant acceleration. However, the calculation would be more complex and would require additional information about the initial vertical velocity.

Q7: What is the relationship between the time of flight and the horizontal distance traveled by the cannonball?

A7: The time of flight is directly proportional to the horizontal distance traveled by the cannonball. This means that if the time of flight is increased, the horizontal distance traveled will also increase.

Conclusion

In conclusion, the motion of a cannonball fired horizontally with a speed of $20.0 , \text{m/s}$ from a height of $68.0 , \text{m}$ is a classic problem in physics. By neglecting air resistance and considering only the force of gravity, we can calculate the time the ball remains in the air. The Q&A section above provides additional information and clarifies some common misconceptions about this problem.

Discussion

The motion of a cannonball fired horizontally is a simple yet fascinating problem that illustrates the principles of motion under gravity. By understanding the relationship between the time of flight and the height from which the cannonball is fired, we can gain a deeper appreciation for the physics of motion.

Key Points

  • The horizontal component of motion remains constant throughout the flight.
  • The acceleration due to gravity affects the time of flight.
  • The time of flight is directly proportional to the square root of the height from which the cannonball is fired.
  • Air resistance is neglected in this problem.
  • The initial vertical velocity is zero.

Further Reading

For further reading on the topic of motion under gravity, we recommend the following resources:

  • [1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics. John Wiley & Sons.
  • [2] Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers. Cengage Learning.

References

[1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics. John Wiley & Sons.

[2] Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers. Cengage Learning.