A Stone Is Thrown Vertically Upward From Ground Level With An Initial Speed Of $v_0 = 20 , \text{m/s}$. The Stone Is Caught By The Thrower At A Height Of $5 , \text{m}$ Above The Ground On Its Way Back.

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Introduction

When a stone is thrown vertically upward from the ground, it experiences a downward acceleration due to gravity, which causes it to decelerate and eventually return to its initial position. In this article, we will explore the physics behind the motion of a stone thrown vertically upward from the ground level with an initial speed of $v_0 = 20 , \text{m/s}$. We will also examine the effect of air resistance and the role of gravity in determining the stone's trajectory.

The Physics of Vertical Motion

When an object is thrown vertically upward, it experiences a downward acceleration due to gravity, which is given by the equation $a = -g$, where $g$ is the acceleration due to gravity, approximately equal to $9.8 , \text{m/s}^2$. This downward acceleration causes the object to decelerate and eventually return to its initial position.

The motion of the stone can be described using the equations of motion, which relate the position, velocity, and acceleration of the object. The position of the stone at any time $t$ is given by the equation $s(t) = s_0 + v_0t - \frac{1}{2}gt^2$, where $s_0$ is the initial position, $v_0$ is the initial velocity, and $g$ is the acceleration due to gravity.

Calculating the Time of Flight

To calculate the time of flight of the stone, we need to determine the time it takes for the stone to reach its maximum height and then return to its initial position. We can do this by setting the position of the stone at the maximum height equal to the initial position and solving for time.

Let's assume that the stone is thrown from the ground level, so the initial position $s_0$ is equal to $0$. The position of the stone at the maximum height is given by the equation $s(t) = v_0t - \frac{1}{2}gt^2$.

Setting the position of the stone at the maximum height equal to the initial position, we get:

v0t−12gt2=0v_0t - \frac{1}{2}gt^2 = 0

Solving for time, we get:

t=2v0gt = \frac{2v_0}{g}

Substituting the values of $v_0$ and $g$, we get:

t=2(20 m/s)9.8 m/s2=4.08 st = \frac{2(20 \, \text{m/s})}{9.8 \, \text{m/s}^2} = 4.08 \, \text{s}

Calculating the Maximum Height

To calculate the maximum height reached by the stone, we need to determine the position of the stone at the maximum height. We can do this by setting the velocity of the stone equal to zero and solving for position.

The velocity of the stone at any time $t$ is given by the equation $v(t) = v_0 - gt$.

Setting the velocity of the stone equal to zero, we get:

v0−gt=0v_0 - gt = 0

Solving for time, we get:

t=v0gt = \frac{v_0}{g}

Substituting the values of $v_0$ and $g$, we get:

t=20 m/s9.8 m/s2=2.04 st = \frac{20 \, \text{m/s}}{9.8 \, \text{m/s}^2} = 2.04 \, \text{s}

Now, we can substitute this value of time into the equation for position to get:

s(t)=v0t−12gt2s(t) = v_0t - \frac{1}{2}gt^2

s(2.04 s)=(20 m/s)(2.04 s)−12(9.8 m/s2)(2.04 s)2s(2.04 \, \text{s}) = (20 \, \text{m/s})(2.04 \, \text{s}) - \frac{1}{2}(9.8 \, \text{m/s}^2)(2.04 \, \text{s})^2

s(2.04 s)=40.8 m−20.8 ms(2.04 \, \text{s}) = 40.8 \, \text{m} - 20.8 \, \text{m}

s(2.04 s)=20 ms(2.04 \, \text{s}) = 20 \, \text{m}

Calculating the Time of Catch

To calculate the time of catch, we need to determine the time it takes for the stone to reach the height of $5 , \text{m}$ above the ground. We can do this by setting the position of the stone equal to $5 , \text{m}$ and solving for time.

The position of the stone at any time $t$ is given by the equation $s(t) = v_0t - \frac{1}{2}gt^2$.

Setting the position of the stone equal to $5 , \text{m}$, we get:

v0t−12gt2=5 mv_0t - \frac{1}{2}gt^2 = 5 \, \text{m}

Solving for time, we get:

t=v0±v02+2g(5 m)gt = \frac{v_0 \pm \sqrt{v_0^2 + 2g(5 \, \text{m})}}{g}

Substituting the values of $v_0$ and $g$, we get:

t=20 m/s±(20 m/s)2+2(9.8 m/s2)(5 m)9.8 m/s2t = \frac{20 \, \text{m/s} \pm \sqrt{(20 \, \text{m/s})^2 + 2(9.8 \, \text{m/s}^2)(5 \, \text{m})}}{9.8 \, \text{m/s}^2}

t=20 m/s±400 m2/s2+98 m2/s29.8 m/s2t = \frac{20 \, \text{m/s} \pm \sqrt{400 \, \text{m}^2/\text{s}^2 + 98 \, \text{m}^2/\text{s}^2}}{9.8 \, \text{m/s}^2}

t=20 m/s±498 m2/s29.8 m/s2t = \frac{20 \, \text{m/s} \pm \sqrt{498 \, \text{m}^2/\text{s}^2}}{9.8 \, \text{m/s}^2}

t=20 m/s±22.3 m/s9.8 m/s2t = \frac{20 \, \text{m/s} \pm 22.3 \, \text{m/s}}{9.8 \, \text{m/s}^2}

We take the positive root, since time cannot be negative:

t=20 m/s+22.3 m/s9.8 m/s2t = \frac{20 \, \text{m/s} + 22.3 \, \text{m/s}}{9.8 \, \text{m/s}^2}

t=42.3 m/s9.8 m/s2t = \frac{42.3 \, \text{m/s}}{9.8 \, \text{m/s}^2}

t=4.31 st = 4.31 \, \text{s}

Conclusion

In this article, we have explored the physics behind the motion of a stone thrown vertically upward from the ground level with an initial speed of $v_0 = 20 , \text{m/s}$. We have calculated the time of flight, the maximum height reached by the stone, and the time of catch. The results show that the stone takes approximately $4.08 , \text{s}$ to reach its maximum height, and approximately $4.31 , \text{s}$ to be caught by the thrower at a height of $5 , \text{m}$ above the ground.

References

  • Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics (10th ed.). John Wiley & Sons.
  • Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers (10th ed.). Cengage Learning.

Glossary

  • Acceleration: The rate of change of velocity with respect to time.
  • Velocity: The rate of change of position with respect to time.
  • Position: The location of an object in space.
  • Time of flight: The time it takes for an object to reach its maximum height and return to its initial position.
  • Maximum height: The highest point reached by an object in its trajectory.
  • Time of catch: The time it takes for an object to be caught by a person or a device.

Introduction

In our previous article, we explored the physics behind the motion of a stone thrown vertically upward from the ground level with an initial speed of $v_0 = 20 , \text{m/s}$. We calculated the time of flight, the maximum height reached by the stone, and the time of catch. In this article, we will answer some of the most frequently asked questions about the motion of a stone thrown vertically upward.

Q: What is the acceleration of the stone?

A: The acceleration of the stone is given by the equation $a = -g$, where $g$ is the acceleration due to gravity, approximately equal to $9.8 , \text{m/s}^2$. This means that the stone is accelerating downward at a rate of $9.8 , \text{m/s}^2$.

Q: What is the velocity of the stone at the maximum height?

A: The velocity of the stone at the maximum height is equal to zero. This is because the stone has reached its maximum height and is momentarily at rest.

Q: What is the position of the stone at the maximum height?

A: The position of the stone at the maximum height is given by the equation $s(t) = v_0t - \frac{1}{2}gt^2$, where $t$ is the time it takes for the stone to reach its maximum height. Substituting the values of $v_0$ and $g$, we get:

s(t)=(20 m/s)(2.04 s)−12(9.8 m/s2)(2.04 s)2s(t) = (20 \, \text{m/s})(2.04 \, \text{s}) - \frac{1}{2}(9.8 \, \text{m/s}^2)(2.04 \, \text{s})^2

s(t)=40.8 m−20.8 ms(t) = 40.8 \, \text{m} - 20.8 \, \text{m}

s(t)=20 ms(t) = 20 \, \text{m}

Q: What is the time of catch?

A: The time of catch is the time it takes for the stone to be caught by the thrower at a height of $5 , \text{m}$ above the ground. We calculated this time to be approximately $4.31 , \text{s}$.

Q: What is the effect of air resistance on the motion of the stone?

A: Air resistance can affect the motion of the stone by slowing it down and causing it to fall faster than it would in the absence of air resistance. However, in this article, we assumed that air resistance is negligible and that the stone is thrown in a vacuum.

Q: Can the stone be caught at a height greater than the maximum height?

A: No, the stone cannot be caught at a height greater than the maximum height. This is because the stone has reached its maximum height and is then falling back down to the ground.

Q: Can the stone be caught at a height less than the maximum height?

A: Yes, the stone can be caught at a height less than the maximum height. This is because the stone is falling back down to the ground and can be caught at any point during its descent.

Q: What is the velocity of the stone at the time of catch?

A: The velocity of the stone at the time of catch is given by the equation $v(t) = v_0 - gt$, where $t$ is the time it takes for the stone to be caught. Substituting the values of $v_0$ and $g$, we get:

v(t)=(20 m/s)−(9.8 m/s2)(4.31 s)v(t) = (20 \, \text{m/s}) - (9.8 \, \text{m/s}^2)(4.31 \, \text{s})

v(t)=20 m/s−42.3 m/sv(t) = 20 \, \text{m/s} - 42.3 \, \text{m/s}

v(t)=−22.3 m/sv(t) = -22.3 \, \text{m/s}

Conclusion

In this article, we have answered some of the most frequently asked questions about the motion of a stone thrown vertically upward. We have discussed the acceleration of the stone, the velocity of the stone at the maximum height, the position of the stone at the maximum height, the time of catch, the effect of air resistance, and the velocity of the stone at the time of catch.

References

  • Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics (10th ed.). John Wiley & Sons.
  • Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers (10th ed.). Cengage Learning.

Glossary

  • Acceleration: The rate of change of velocity with respect to time.
  • Velocity: The rate of change of position with respect to time.
  • Position: The location of an object in space.
  • Time of flight: The time it takes for an object to reach its maximum height and return to its initial position.
  • Maximum height: The highest point reached by an object in its trajectory.
  • Time of catch: The time it takes for an object to be caught by a person or a device.