A Particle Is Projected Vertically Upwards From A Point O With A Speed Of $40 , \text{m/s}$. I. Find The Maximum Height Reached.II. Find The Times When The Particle Is 30 M Above Point O, Correct To The Nearest Whole Number.

Mar 9, 2025 by ADMIN 227 views

A Particle Projected Vertically Upwards: Maximum Height and Time of Flight

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Introduction

In this article, we will explore the motion of a particle projected vertically upwards from a point O with an initial speed of 40 m/s. We will use the principles of kinematics to find the maximum height reached by the particle and the times when the particle is 30 m above point O.

I. Maximum Height Reached

To find the maximum height reached by the particle, we can use the equation of motion under gravity:

s = ut + \frac{1}{2}gt^2

where $s$ is the displacement, $u$ is the initial velocity, $t$ is the time, and $g$ is the acceleration due to gravity (approximately 9.8 m/s^2).

Since the particle is projected vertically upwards, the displacement $s$ will be maximum when the velocity of the particle becomes zero. At this point, the particle will be at its maximum height.

Let's denote the maximum height as $h$ . Then, we can write:

h = ut + \frac{1}{2}gt^2

Since the particle is projected upwards, the initial velocity $u$ is positive. We can also assume that the time taken to reach the maximum height is $t_m$ . Then, we can write:

h = ut_m + \frac{1}{2}gt_m^2

Substituting the values of $u$ , $g$ , and $t_m$ , we get:

h = 40t_m + \frac{1}{2} \times 9.8 \times t_m^2

Simplifying the equation, we get:

h = 40t_m + 4.9t_m^2

To find the maximum height, we need to find the value of $t_m$ . We can do this by setting the velocity of the particle to zero:

v = u - gt = 0

Substituting the values of $u$ and $g$ , we get:

40 - 9.8t = 0

Solving for $t$ , we get:

t_m = \frac{40}{9.8} = 4.08 \, \text{s}

Substituting the value of $t_m$ into the equation for $h$ , we get:

h = 40 \times 4.08 + 4.9 \times (4.08)^2

Simplifying the equation, we get:

h = 163.2 + 67.3 = 230.5 \, \text{m}

Therefore, the maximum height reached by the particle is approximately 230.5 m.

II. Times When the Particle is 30 m Above Point O

To find the times when the particle is 30 m above point O, we can use the equation of motion under gravity:

s = ut + \frac{1}{2}gt^2

We can set $s$ to 30 m and solve for $t$ :

30 = 40t + \frac{1}{2} \times 9.8 \times t^2

Simplifying the equation, we get:

30 = 40t + 4.9t^2

Rearranging the equation, we get:

4.9t^2 + 40t - 30 = 0

Solving the quadratic equation, we get:

t = \frac{-40 \pm \sqrt{40^2 + 4 \times 4.9 \times 30}}{2 \times 4.9}

Simplifying the equation, we get:

t = \frac{-40 \pm \sqrt{1600 + 588}}{9.8}

t = \frac{-40 \pm \sqrt{2188}}{9.8}

t = \frac{-40 \pm 46.8}{9.8}

Solving for $t$ , we get two possible values:

t_1 = \frac{-40 + 46.8}{9.8} = 0.53 \, \text{s}

t_2 = \frac{-40 - 46.8}{9.8} = -5.33 \, \text{s}

Since time cannot be negative, we discard the second solution. Therefore, the particle is 30 m above point O at $t = 0.53 \, \text{s}$ .

Conclusion

In this article, we have used the principles of kinematics to find the maximum height reached by a particle projected vertically upwards from a point O with an initial speed of 40 m/s. We have also found the times when the particle is 30 m above point O. The results show that the maximum height reached by the particle is approximately 230.5 m, and the particle is 30 m above point O at $t = 0.53 \, \text{s}$ .

Discussion

The results of this article can be used to understand the motion of objects under gravity. The equation of motion under gravity can be used to find the displacement, velocity, and acceleration of an object at any given time. The results of this article can also be used to design and analyze the motion of objects in various fields, such as engineering and physics.

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.

Note: The references provided are for general information purposes only and are not directly related to the specific problem discussed in this article.

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Introduction

In our previous article, we explored the motion of a particle projected vertically upwards from a point O with an initial speed of 40 m/s. We used the principles of kinematics to find the maximum height reached by the particle and the times when the particle is 30 m above point O. In this article, we will answer some frequently asked questions related to the motion of the particle.

Q&A

Q1: What is the initial velocity of the particle?

A1: The initial velocity of the particle is 40 m/s.

Q2: What is the maximum height reached by the particle?

A2: The maximum height reached by the particle is approximately 230.5 m.

Q3: At what time is the particle 30 m above point O?

A3: The particle is 30 m above point O at $t = 0.53 \, \text{s}$ .

Q4: What is the acceleration due to gravity?

A4: The acceleration due to gravity is approximately 9.8 m/s^2.

Q5: How can we find the displacement of the particle at any given time?

A5: We can use the equation of motion under gravity:

s = ut + \frac{1}{2}gt^2

where $s$ is the displacement, $u$ is the initial velocity, $t$ is the time, and $g$ is the acceleration due to gravity.

Q6: What is the velocity of the particle at the maximum height?

A6: The velocity of the particle at the maximum height is zero.

Q7: How can we find the time taken to reach the maximum height?

A7: We can use the equation:

v = u - gt = 0

where $v$ is the velocity, $u$ is the initial velocity, $g$ is the acceleration due to gravity, and $t$ is the time.

Q8: What is the significance of the negative solution in the quadratic equation?

A8: The negative solution in the quadratic equation is not physically meaningful, as time cannot be negative.

Q9: How can we use the results of this article in real-world applications?

A9: The results of this article can be used to design and analyze the motion of objects in various fields, such as engineering and physics.

Q10: What are some common mistakes to avoid when solving problems related to motion under gravity?

A10: Some common mistakes to avoid when solving problems related to motion under gravity include:

Not considering the direction of the acceleration due to gravity
Not using the correct equation of motion
Not checking the units of the variables
Not considering the limitations of the model

Conclusion

In this article, we have answered some frequently asked questions related to the motion of a particle projected vertically upwards from a point O with an initial speed of 40 m/s. We hope that this article has provided a clear understanding of the principles of kinematics and the motion of objects under gravity.

Discussion

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.

Note: The references provided are for general information purposes only and are not directly related to the specific problem discussed in this article.