Chapter 2, Problem 79P

(a)

To determine

Whether the rocket designed to be used to sample the local atmosphere for pollution, achieves its goal of reaching 20 km.

Answer to Problem 79P

The rocket reaches a height of $18.95\text{ km}$ and therefore, does not achieve its goal. To reach a height of $\text{20 km}$ (i) the initial acceleration should be increased or (ii) the time during which the rocket accelerates should be increased.

Explanation of Solution

Given:

The initial velocity of the rocket

  $v_{0y}=0\text{ m/s}$

The magnitude of its upward acceleration

  $a_y=20{\text{ m/s}}^2$

Time during which the rocket accelerates upwards

  $t_1=25\text{ s}$

The height the rocket should reach

  $h=20\text{ km}$

Formula used:

The rocket accelerates upwards for a time $t_1$ .

Assume a sign convention where the $+y$ is directed upwards and the $-y$ is directed downwards.

The distance travelled by the rocket during this time period is given by the following expression:

  $\Delta y_1=v_{0y}{t}_1+\frac{1}{2}{a}_y{t}_1^2$  ........(1)

At the end of the time interval $t_1$ , the rocket’s velocity is calculated using the expression,

  $v_y^2=v_{1y}^2+2g\Delta y_2$  .........(2)

After a time $t_1$ , the engines are switched off, hence the rocket moves under the action of the gravitational force. It continues to move upward, since its velocity is directed upwards. The velocity of the rocket reduces since the gravitational force acts opposite to the direction of the rocket’s velocity and reaches a value $v_y=0\text{ m/s}$ in a time $t_2$ . During this time, the rocket travels a vertical distance of $\Delta y_2$ , which can be calculated using the following equation:

  $v_y^2=v_{0y}^2+2g\Delta y_2$  ........(3)

Here, $g$ is the acceleration of free fall and has a value $g=-9.81{\text{ m/s}}^2$ . The negative sign shows that $g$ is directed downwards along the $-y$ direction.

The total height travelled by the rocket is the sum of the two distances $\Delta y_1$ and $\Delta y_2$ .

  $\Delta y=\Delta y_1+\Delta y_2$  ........(4)

Calculation:

Calculate the vertical distance $\Delta y_1$ travelled by the rocket when its engine is on and it accelerates upwards by substituting the values of the variables in equation (1).

  $\begin{array}{c}\Delta y_1=v_{0y}{t}_1+\frac{1}{2}{a}_y{t}_1^2\\ =\left(0\text{ m/s}\right)\left(25\text{ s}\right)+\frac{1}{2}\left(20{\text{ m/s}}^2\right){\left(25\text{ s}\right)}^2\\ =6.25\times {10}^3\text{ m}\\ =6.25\text{ km}\end{array}$

Using equation (2), calculate the value of the rocket’s speed $v_{1y}$ at the instant when its engine is switched off.

  $\begin{array}{c}{v}_{1y}=v_{0y}+a_y{t}_1\\ =\left(0\text{ m/s}\right)+\left(20{\text{ m/s}}^2\right)\left(25\text{ s}\right)\\ =500\text{ m/s}\end{array}$

After the engines are switched off, the rocket decelerates due to the action of the acceleration of free fall. When it reaches the maximum point in its trajectory, its final velocity becomes zero.

Substitute $\left(0\text{ m/s}\right)$ for $v_y$ , $\left(500\text{ m/s}\right)$ for $v_{1y}$ and $\left(-9.81{\text{ m/s}}^2\right)$ for $g$ in equation (3) and calculate the distance travelled by the rocket after the engines were switched off.

  $\begin{array}{c}{v}_y^2=v_{1y}^2+2g\Delta y_2\\ \left(0\text{ m/s}\right)=\left(500\text{ m/s}\right)+2\left(-9.81{\text{ m/s}}^2\right)\Delta y_2\end{array}$

Therefore,

  $\begin{array}{c}\Delta y_2=\frac{\left(500\text{ m/s}\right)}{2\left(9.81{\text{ m/s}}^2\right)}\\ =1.27\times {10}^4\text{m}\\ =12.7\text{ km}\end{array}$

Calculate the maximum vertical distance travelled by the rocket by substituting the calculated values of $\Delta y_1$ and $\Delta y_2$ in equation (4).

  $\begin{array}{c}\Delta y=\Delta y_1+\Delta y_2\\ =\left(6.25\text{ km}\right)+\left(12.7\text{ km}\right)\\ =18.95\text{ km}\end{array}$

Conclusion:

The rocket was designed to sample air at a height of $20\text{ km}$ to test for pollution. But it can be seen that the rocket reaches only a vertical height of $18.95\text{ km}$ . Hence, it does not achieve its goal.

For the rocket to reach $20\text{ km}$ , the value of the upwards acceleration provided by the engines could be increased. This would give a larger speed to the rocket when the engine switches off and also increase the distance travelled during acceleration. If it is not possible to increase the value of acceleration, the engine can run for more than $25\text{ s}$ . Both these methods would result in the increase in the values of both $\Delta y_1$ and $\Delta y_2$ , hence increase the total vertical distance.

(b)

To determine

The total time the rocket is in air.

Answer to Problem 79P

The rocket is in air for a total time of $\text{138 s}$ .

Explanation of Solution

Given:

The velocity of the rocket when engine is switched off

  $v_{1y}=500\text{ m/s}$

The total vertical distance travelled by the rocket.

  $\Delta y=1.895\times {10}^4\text{m}$

Time during which the rocket accelerates upwards

  $t_1=25\text{ s}$

Velocity at the point of maximum height

  $v_y=0\text{ m/s}$

Formula used:

The time taken by the rocket to reach its maximum height after the engine is switched off is calculated using the expression:

  $v_y=v_{1y}+g{t}_2$  ........(5)

When the rocket reaches the point of maximum height its velocity becomes zero and it starts to fall down.

The time taken by the rocket to fall is given by the following expression:

  $-\Delta y=v_y{t}_3+\frac{1}{2}g{t}_3^2$  ........(6)

The negative sign shows that the displacement is made in the downward direction.

The total time the rocket is in air is the sum of (i) the time taken by it to accelerate upwards(ii) time taken to reach the maximum height after the engine is switched off and (iii) time taken to fall to the ground from the point of maximum height.

Therefore,

  $t=t_1+t_2+t_3$  ........(7)

Calculation:

Substitute the given values of variables in equation (5) and calculate the time $t_2$ taken by the rocket to reach the maximum height after the engine is switched off.

  $\begin{array}{c}{v}_y=v_{1y}+g{t}_2\\ \left(0\text{ m/s}\right)=\left(500\text{ m/s}\right)+\left(-9.81{\text{ m/s}}^2\right)t_2\\ t_2=\frac{\left(500\text{ m/s}\right)}{\left(9.81{\text{ m/s}}^2\right)}\\ =51\text{ s}\end{array}$

Substitute the values of variables in equation (6) and calculate the time $t_3$ taken by the rocket to reach the ground.

  $\begin{array}{c}-\Delta y=v_y{t}_3+\frac{1}{2}g{t}_3^2\\ \left(-1.895\times {10}^4\text{m}\right)=\left(0\text{ m/s}\right)t_3+\frac{1}{2}\left(-9.81{\text{ m/s}}^2\right)t_3^2\\ t_3=\sqrt{\frac{2\left(1.895\times {10}^4\text{m}\right)}{\left(9.81{\text{ m/s}}^2\right)}}\\ =62.2\text{ s}\end{array}$

Substitute the values of $t_1$ , $t_2$ and $t_3$ in equation (7) and calculate the total time the rocket was in air.

  $\begin{array}{c}t=t_1+t_2+t_3\\ =\left(25\text{ s}\right)+\left(51\text{ s}\right)+\left(62.2\text{ s}\right)\\ =138.2\text{ s}\\ \text{=138 s}\end{array}$

Conclusion:

Thus, the rocket is in air for a total time of $\text{138 s}$ .

(c)

To determine

To determine the speed of the rocket just before it hits the ground.

Answer to Problem 79P

The speed of the rocket just before it hits the ground is found to be $\text{610 m/s}$ .

Explanation of Solution

Given:

The vertical distance the rocket falls from the point of maximum height.

  $\Delta y=1.895\times {10}^4\text{m}$

Velocity at the point of maximum height

  $v_y=0\text{ m/s}$

Time taken by the rocket to fall to the ground

  $t_3=62.2\text{ s}$

Formula used:

The speed of the rocket when it just reaches the ground is calculated using the following expression:

  $v=v_y+g{t}_3$  ........(8)

Calculation:

Substitute the values of the variables in equation (8) and calculate the rocket’s speed when it hits the ground.

  $\begin{array}{c}v=v_y+g{t}_3\\ =\left(0\text{ m/s}\right)+\left(-9.81{\text{ m/s}}^2\right)\left(62.2\text{ s}\right)\\ =-610\text{ m/s}\end{array}$

The negative sign shows that its velocity is directed downwards along the −y direction.

Conclusion:

Thus, the speed of the rocket just before it hits the ground is found to be $\text{610 m/s}$ .