If A 45 Kg Object Is Dropped From A Height Of 6.0 M, What Will Be Its Velocity When It Is Halfway Toward The Ground? (Use $g=9.80 , \text{m/s}^2$ And Ignore Air Resistance.)A. 7.7 M/s 7.7 \, \text{m/s} 7.7 M/s B. 11 M/s 11 \, \text{m/s} 11 M/s C.

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Introduction

When an object is dropped from a certain height, it accelerates downward due to the force of gravity. The acceleration due to gravity is denoted by the symbol $g$ and is approximately equal to $9.80 , \text{m/s}^2$ on the surface of the Earth. In this article, we will explore the physics of a falling object and determine its velocity when it is halfway toward the ground.

The Physics of Falling Objects

When an object is dropped from a height, it experiences a downward force due to gravity. This force causes the object to accelerate downward, and its velocity increases as it falls. The acceleration due to gravity is constant and is given by the equation:

a=g=9.80 m/s2a = g = 9.80 \, \text{m/s}^2

The velocity of the object can be calculated using the equation:

v=u+atv = u + at

where $v$ is the final velocity, $u$ is the initial velocity (which is zero in this case), $a$ is the acceleration, and $t$ is the time.

Calculating the Velocity of the Object

In this problem, we are given that the object is dropped from a height of $6.0 , \text{m}$. We need to calculate its velocity when it is halfway toward the ground. To do this, we need to calculate the time it takes for the object to fall halfway.

The distance traveled by the object is given by the equation:

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

Since the object is dropped from rest, the initial velocity $u$ is zero. Therefore, the equation becomes:

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

We know that the object falls a distance of $3.0 , \text{m}$ (half of the total height) in the time $t$. Therefore, we can write:

3.0=12×9.80×t23.0 = \frac{1}{2} \times 9.80 \times t^2

Solving for $t$, we get:

t=2×3.09.80=0.67 st = \sqrt{\frac{2 \times 3.0}{9.80}} = 0.67 \, \text{s}

Now that we have the time, we can calculate the velocity of the object using the equation:

v=u+at=0+9.80×0.67=6.57 m/sv = u + at = 0 + 9.80 \times 0.67 = 6.57 \, \text{m/s}

However, this is not the correct answer. We need to consider the fact that the object is accelerating downward, and its velocity is increasing as it falls. Therefore, we need to use the equation:

v=2asv = \sqrt{2as}

where $v$ is the final velocity, $a$ is the acceleration, and $s$ is the distance traveled.

Substituting the values, we get:

v=2×9.80×3.0=7.7 m/sv = \sqrt{2 \times 9.80 \times 3.0} = 7.7 \, \text{m/s}

Conclusion

In this article, we explored the physics of a falling object and determined its velocity when it is halfway toward the ground. We used the equations of motion to calculate the time it takes for the object to fall halfway and then used the equation for velocity to calculate the final velocity. The correct answer is $7.7 , \text{m/s}$.

Discussion

This problem is a classic example of a falling object under the influence of gravity. The acceleration due to gravity is a fundamental concept in physics, and it is essential to understand how it affects the motion of objects. The equations of motion are powerful tools that can be used to solve problems involving falling objects.

References

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

Additional Resources

Q: What is the acceleration due to gravity?

A: The acceleration due to gravity is a fundamental concept in physics and is denoted by the symbol $g$. It is approximately equal to $9.80 , \text{m/s}^2$ on the surface of the Earth.

Q: What is the equation for the velocity of a falling object?

A: The equation for the velocity of a falling object is:

v=u+atv = u + at

where $v$ is the final velocity, $u$ is the initial velocity (which is zero in this case), $a$ is the acceleration, and $t$ is the time.

Q: How do I calculate the time it takes for an object to fall a certain distance?

A: To calculate the time it takes for an object to fall a certain distance, you can use the equation:

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

Since the object is dropped from rest, the initial velocity $u$ is zero. Therefore, the equation becomes:

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

You can then solve for $t$.

Q: What is the equation for the velocity of a falling object when it is halfway toward the ground?

A: The equation for the velocity of a falling object when it is halfway toward the ground is:

v=2asv = \sqrt{2as}

where $v$ is the final velocity, $a$ is the acceleration, and $s$ is the distance traveled.

Q: Can I use the equation $v = u + at$ to calculate the velocity of a falling object when it is halfway toward the ground?

A: No, you cannot use the equation $v = u + at$ to calculate the velocity of a falling object when it is halfway toward the ground. This equation is only valid for objects that are falling from rest, and it does not take into account the fact that the object is accelerating downward.

Q: What is the difference between the acceleration due to gravity and the velocity of a falling object?

A: The acceleration due to gravity is a constant value that is approximately equal to $9.80 , \text{m/s}^2$ on the surface of the Earth. The velocity of a falling object, on the other hand, is a variable that depends on the time and distance traveled by the object.

Q: Can I use the equation $v = \sqrt{2as}$ to calculate the velocity of a falling object at any point in its trajectory?

A: Yes, you can use the equation $v = \sqrt{2as}$ to calculate the velocity of a falling object at any point in its trajectory. This equation is valid for objects that are falling under the influence of gravity, and it takes into account the fact that the object is accelerating downward.

Q: What is the significance of the equation $v = \sqrt{2as}$ in the context of falling objects?

A: The equation $v = \sqrt{2as}$ is a fundamental equation in the context of falling objects. It describes the relationship between the velocity of a falling object and the distance it has traveled, and it is a key concept in the study of motion under the influence of gravity.

Q: Can I use the equation $v = u + at$ to calculate the velocity of a falling object at any point in its trajectory?

A: No, you cannot use the equation $v = u + at$ to calculate the velocity of a falling object at any point in its trajectory. This equation is only valid for objects that are falling from rest, and it does not take into account the fact that the object is accelerating downward.

Q: What is the difference between the equation $v = u + at$ and the equation $v = \sqrt{2as}$?

A: The equation $v = u + at$ is a simple equation that describes the relationship between the velocity of an object and the time it has traveled. The equation $v = \sqrt{2as}$, on the other hand, is a more complex equation that describes the relationship between the velocity of a falling object and the distance it has traveled.