4. An Aid Parcel Is Released From A Plane Flying Horizontally At A Velocity Of $60 \, \text{m/s}$. It Is At A Height Of $1000 \, \text{m}$. (Use $g = 9.8 \, \text{m/s}^2$.)i. Write Down The 3 Equations For Downward Motion

Introduction

In this problem, we are tasked with analyzing the motion of an aid parcel released from a plane flying horizontally at a velocity of $60 \, \text{m/s}$. The parcel is initially at a height of $1000 \, \text{m}$ above the ground. We will use the equations of motion to describe the downward motion of the parcel.

Given Information

• Initial velocity ($v_0$): $60 \, \text{m/s}$
• Initial height ($h_0$): $1000 \, \text{m}$
• Acceleration due to gravity ($g$): $9.8 \, \text{m/s}^2$
• Time ($t$): unknown

Equations of Motion

The equations of motion for an object under constant acceleration are:

1. Equation of motion for position: $s(t) = s_0 + v_0t + \frac{1}{2}at^2$

In this case, the position of the parcel is its height above the ground, which is a function of time.

2. Equation of motion for velocity: $v(t) = v_0 + at$

The velocity of the parcel is its rate of change of position with respect to time.

3. Equation of motion for acceleration: $a = \text{constant}$

In this case, the acceleration is due to gravity, which is a constant $9.8 \, \text{m/s}^2$.

Applying the Equations

We can now apply these equations to the problem at hand.

Equation 1: Position as a Function of Time

We start by substituting the given values into the equation of motion for position:

$s(t) = s_0 + v_0t + \frac{1}{2}at^2$

$s(t) = 1000 + 60t - \frac{1}{2} \cdot 9.8t^2$

This equation describes the height of the parcel above the ground as a function of time.

Equation 2: Velocity as a Function of Time

Next, we substitute the given values into the equation of motion for velocity:

$v(t) = v_0 + at$

$v(t) = 60 - 9.8t$

This equation describes the velocity of the parcel as a function of time.

Equation 3: Acceleration as a Constant

Finally, we note that the acceleration is a constant $9.8 \, \text{m/s}^2$, which is due to gravity.

Discussion

In this problem, we have derived the three equations of motion for an aid parcel released from a plane flying horizontally. We have applied these equations to the problem at hand, obtaining expressions for the position, velocity, and acceleration of the parcel as functions of time.

The position of the parcel is described by the equation $s(t) = 1000 + 60t - \frac{1}{2} \cdot 9.8t^2$, which shows that the parcel's height above the ground decreases quadratically with time.

The velocity of the parcel is described by the equation $v(t) = 60 - 9.8t$, which shows that the parcel's velocity decreases linearly with time.

The acceleration of the parcel is a constant $9.8 \, \text{m/s}^2$, which is due to gravity.

Conclusion

In conclusion, we have analyzed the motion of an aid parcel released from a plane flying horizontally. We have derived the three equations of motion for the parcel and applied them to the problem at hand. The results show that the parcel's height above the ground decreases quadratically with time, its velocity decreases linearly with time, and its acceleration is a constant due to gravity.

Further Questions

1. What is the time it takes for the parcel to reach the ground?
2. What is the velocity of the parcel when it reaches the ground?
3. What is the acceleration of the parcel when it reaches the ground?

Introduction

In our previous article, we analyzed the motion of an aid parcel released from a plane flying horizontally. We derived the three equations of motion for the parcel and applied them to the problem at hand. In this article, we will answer some of the questions that were left unanswered.

Q1: What is the time it takes for the parcel to reach the ground?

To find the time it takes for the parcel to reach the ground, we need to set the position equation equal to zero and solve for time:

$s(t) = 1000 + 60t - \frac{1}{2} \cdot 9.8t^2 = 0$

We can rearrange this equation to get a quadratic equation in terms of time:

$4.9t^2 - 60t - 1000 = 0$

We can solve this equation using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 4.9$, $b = -60$, and $c = -1000$.

Plugging in these values, we get:

$t = \frac{60 \pm \sqrt{(-60)^2 - 4(4.9)(-1000)}}{2(4.9)}$

Simplifying this expression, we get:

$t = \frac{60 \pm \sqrt{3600 + 19600}}{9.8}$

$t = \frac{60 \pm \sqrt{24600}}{9.8}$

$t = \frac{60 \pm 157.1}{9.8}$

We have two possible values for time:

$t_1 = \frac{60 + 157.1}{9.8} = 19.3 \, \text{s}$

$t_2 = \frac{60 - 157.1}{9.8} = -9.5 \, \text{s}$

Since time cannot be negative, we discard the second solution and conclude that the parcel reaches the ground in $19.3 \, \text{s}$.

Q2: What is the velocity of the parcel when it reaches the ground?

To find the velocity of the parcel when it reaches the ground, we need to plug in the value of time we found in the previous question into the velocity equation:

$v(t) = 60 - 9.8t$

$v(19.3) = 60 - 9.8(19.3)$

$v(19.3) = 60 - 189.14$

$v(19.3) = -129.14 \, \text{m/s}$

So, the velocity of the parcel when it reaches the ground is $-129.14 \, \text{m/s}$.

Q3: What is the acceleration of the parcel when it reaches the ground?

The acceleration of the parcel is a constant $9.8 \, \text{m/s}^2$, which is due to gravity. Therefore, the acceleration of the parcel when it reaches the ground is still $9.8 \, \text{m/s}^2$.

Conclusion

In this article, we answered some of the questions that were left unanswered in our previous article. We found that the parcel reaches the ground in $19.3 \, \text{s}$, its velocity when it reaches the ground is $-129.14 \, \text{m/s}$, and its acceleration when it reaches the ground is still $9.8 \, \text{m/s}^2$.

Further Questions

1. What is the maximum height reached by the parcel?
2. What is the time it takes for the parcel to reach its maximum height?
3. What is the velocity of the parcel at its maximum height?

These questions can be answered by analyzing the equations of motion for the parcel.