A Ball Is Thrown Directly Upward From The Ground With An Initial Velocity Of $4.8 , \text{ft/sec}$. Represent The Height Of The Ball From The Ground $t$ Seconds After It Was Thrown Upward Using The Model

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Introduction

In physics, projectiles are objects that are thrown or launched into the air and follow a curved trajectory under the influence of gravity. One of the fundamental concepts in projectile motion is the height of the object as a function of time. In this article, we will explore the height of a ball thrown directly upward from the ground with an initial velocity of $4.8 , \text{ft/sec}$ using the model $h(t) = h_0 + v_0t - \frac{1}{2}gt^2$, where $h(t)$ is the height of the ball at time $t$, $h_0$ is the initial height (which is 0 in this case), $v_0$ is the initial velocity, and $g$ is the acceleration due to gravity.

The Model

The model for the height of a projectile is given by the equation $h(t) = h_0 + v_0t - \frac{1}{2}gt^2$. In this case, the initial height $h_0$ is 0, since the ball is thrown from the ground. The initial velocity $v_0$ is $4.8 , \text{ft/sec}$, and the acceleration due to gravity $g$ is $32 , \text{ft/sec}^2$. Plugging these values into the equation, we get:

h(t)=0+4.8t−12(32)t2h(t) = 0 + 4.8t - \frac{1}{2}(32)t^2

Simplifying the equation, we get:

h(t)=4.8t−16t2h(t) = 4.8t - 16t^2

Understanding the Model

The model $h(t) = 4.8t - 16t^2$ represents the height of the ball as a function of time. The first term, $4.8t$, represents the initial velocity of the ball, which is $4.8 , \text{ft/sec}$. The second term, $-16t^2$, represents the acceleration due to gravity, which is $-32 , \text{ft/sec}^2$. The negative sign indicates that the acceleration is in the opposite direction of the initial velocity.

Graphing the Model

To visualize the height of the ball as a function of time, we can graph the model $h(t) = 4.8t - 16t^2$. The graph will be a parabola that opens downward, since the acceleration due to gravity is negative.

Finding the Maximum Height

The maximum height of the ball occurs when the velocity of the ball is zero. To find the maximum height, we need to find the time at which the velocity is zero. The velocity of the ball is given by the derivative of the height function:

v(t)=dhdt=4.8−32tv(t) = \frac{dh}{dt} = 4.8 - 32t

Setting the velocity equal to zero, we get:

4.8−32t=04.8 - 32t = 0

Solving for $t$, we get:

t=4.832=0.15 sect = \frac{4.8}{32} = 0.15 \, \text{sec}

Substituting this value of $t$ into the height function, we get:

h(0.15)=4.8(0.15)−16(0.15)2=0.72 fth(0.15) = 4.8(0.15) - 16(0.15)^2 = 0.72 \, \text{ft}

Therefore, the maximum height of the ball is $0.72 , \text{ft}$, which occurs $0.15 , \text{sec}$ after the ball is thrown.

Conclusion

In this article, we have explored the height of a ball thrown directly upward from the ground with an initial velocity of $4.8 , \text{ft/sec}$. We have used the model $h(t) = h_0 + v_0t - \frac{1}{2}gt^2$ to represent the height of the ball as a function of time. We have also graphed the model and found the maximum height of the ball, which occurs $0.15 , \text{sec}$ after the ball is thrown.

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.

Additional Resources

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Frequently Asked Questions

Q: What is the initial velocity of the ball?

A: The initial velocity of the ball is $4.8 , \text{ft/sec}$.

Q: What is the acceleration due to gravity?

A: The acceleration due to gravity is $32 , \text{ft/sec}^2$.

Q: What is the maximum height of the ball?

A: The maximum height of the ball is $0.72 , \text{ft}$, which occurs $0.15 , \text{sec}$ after the ball is thrown.

Q: How long does it take for the ball to reach its maximum height?

A: It takes $0.15 , \text{sec}$ for the ball to reach its maximum height.

Q: What is the velocity of the ball at its maximum height?

A: The velocity of the ball at its maximum height is zero.

Q: What is the height of the ball at $t = 0$?

A: The height of the ball at $t = 0$ is $0 , \text{ft}$, since the ball is thrown from the ground.

Q: What is the height of the ball at $t = 1$?

A: To find the height of the ball at $t = 1$, we need to plug $t = 1$ into the height function:

h(1)=4.8(1)−16(1)2=−11.2 fth(1) = 4.8(1) - 16(1)^2 = -11.2 \, \text{ft}

Therefore, the height of the ball at $t = 1$ is $-11.2 , \text{ft}$.

Q: What is the height of the ball at $t = 2$?

A: To find the height of the ball at $t = 2$, we need to plug $t = 2$ into the height function:

h(2)=4.8(2)−16(2)2=−31.2 fth(2) = 4.8(2) - 16(2)^2 = -31.2 \, \text{ft}

Therefore, the height of the ball at $t = 2$ is $-31.2 , \text{ft}$.

Q: What is the height of the ball at $t = 3$?

A: To find the height of the ball at $t = 3$, we need to plug $t = 3$ into the height function:

h(3)=4.8(3)−16(3)2=−50.4 fth(3) = 4.8(3) - 16(3)^2 = -50.4 \, \text{ft}

Therefore, the height of the ball at $t = 3$ is $-50.4 , \text{ft}$.

Additional Questions

Q: What is the relationship between the height of the ball and the time it takes to reach the ground?

A: The height of the ball is a quadratic function of time, which means that it will reach its maximum height and then decrease back to zero as time increases.

Q: How does the initial velocity affect the height of the ball?

A: The initial velocity affects the height of the ball by determining the rate at which the ball rises. A higher initial velocity will result in a higher maximum height.

Q: How does the acceleration due to gravity affect the height of the ball?

A: The acceleration due to gravity affects the height of the ball by determining the rate at which the ball falls. A higher acceleration due to gravity will result in a lower maximum height.

Conclusion

In this article, we have answered some frequently asked questions about the height of a ball thrown directly upward from the ground with an initial velocity of $4.8 , \text{ft/sec}$. We have also explored the relationship between the height of the ball and the time it takes to reach the ground, as well as the effects of the initial velocity and acceleration due to gravity on the height of the ball.

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.

Additional Resources