Chapter 2, Problem 53P

A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates with a constant upward acceleration of 2.00 m/s² until its engines stop at an altitude of 150. m. (a) What can you say about, the motion of the rocket alter its engines stop? (b) What is the maximum height reached by the rocket? (c) How long after liftoff does the rocket reach its maximum height? (d) How long is the rocket in the air?

To determine

The motion of the rocket after the engines stop.

Answer to Problem 53P

After its engines stop, the rocket is a freely falling body under gravity. It continues upward and eventually slows under the influence of gravity. The rocket comes to rest momentarily at its maximum altitude. Then it falls back to Earth, gaining speed as it falls due to the acceleration due to gravity.

Explanation of Solution

The motion of the rocket can be determined using the acceleration of the rocket.

When the rocket moves upwards, then the engines forces the rocket to move upwards. When there is no external force acting on the rocket, the rocket is moving under the gravitational force. The magnitude of the acceleration acting on the rocket is the acceleration due to gravity. The acceleration due to gravity acts always downwards.

When the engine stops, the rocket starts moves under the acceleration due to gravity and which is opposite to the direction of the motion. This will slow down the rocket. As the rocket reaches its maximum height, the rocket has zero velocity. Then it falls under gravity. The acceleration and the motion of the rocket go on the same direction. This results in speeding up the rocket from zero to the maximum speed when it reaches the ground.

Conclusion:

After its engines stop, the rocket is a freely falling body under gravity. It continues upward and eventually slows under the influence of gravity. The rocket comes to rest momentarily at its maximum altitude. Then it falls back to Earth, gaining speed as it falls due to the acceleration due to gravity.

To determine

The maximum height the rocket reaches.

Answer to Problem 53P

The maximum height the rocket reaches is $308\text{m}$.

Explanation of Solution

Given info: The initial velocity of the of the rocket is $50.0\text{m/s}$, the upward acceleration of the of the rocket is $2.00{\text{m/s}}^2$, the height at which the engine stops is $150\text{m}$, the velocity of the rocket at maximum height is $0$, and the acceleration of the rocket after the engine stops is $-9.80{\text{m/s}}^2$.

Explanation:

The formula used to calculate the final velocity of the rocket when it is accelerating upwards is,

$v=\sqrt{v_0^2+2a\Delta y}$

Here,

$\left(\Delta y\right)$ is the height at which the engine stops

$v_0$ is the initial velocity of the rocket

$v_f$ is the final velocity of the rocket at maximum height

$a$ is the upward acceleration

Substitute $50.0\text{m/s}$ for $v_0$, $2.00{\text{m/s}}^2$ for $a$ and $150\text{m}$ for $\Delta y$ to find $v$.

$\begin{array}{c}v=\sqrt{{\left(50.0\text{m/s}\right)}^2+2\left(2.00{\text{m/s}}^2\right)\left(150\text{m}\right)}\\ =55.7\text{m/s}\end{array}$

Thus, The final velocity of the rocket when the engine stops is $55.7\text{m/s}$.

The formula used to calculate the displacement of the rocket after the engine stops is,

$\Delta y\text{'}=\frac{v_f^2-v^2}{2a\text{'}}$

Here,

$\Delta y\text{'}$ is the displacement of the rocket after the engine stops

$v_f$ is the final velocity of the rocket at maximum height

$a\text{'}$ is the acceleration of the rocket after the engine stops

Substitute $0$ for $v_f$, $55.7\text{m/s}$ for $v$ and $-9.80{\text{m/s}}^2$ for $a$ to find $\Delta y\text{'}$.

$\begin{array}{c}\Delta y\text{'}=\frac{{\left(0\right)}^2-{\left(55.7\text{m/s}\right)}^2}{2\left(-9.80{\text{m/s}}^2\right)}\\ =158\text{m}\end{array}$

Thus, The displacement of the rocket after the engine stops is $158\text{m}$.

The formula used to calculate the maximum height the rocket reaches is,

$h_{\max }=\Delta y+\Delta y\text{'}$

Here,

$h_{\max }$ is the maximum height the rocket reaches

Substitute $150\text{m}$ for $\Delta y$ and $158\text{m}$ for $\Delta y\text{'}$ to find $h_{\max }$.

$\begin{array}{c}{h}_{\max }=150\text{m}+158\text{m}\\ =\text{308}\text{m}\end{array}$

Thus, the maximum height the rocket reaches is $308\text{m}$.

Conclusion:

The maximum height the rocket reaches is $308\text{m}$.

To determine

The time taken by the rocket to reach the maximum height.

Answer to Problem 53P

The time taken by the rocket to reach the maximum height $8.51\text{s}$.

Explanation of Solution

The formula used to calculate the interval at which the rocket moves with an upward acceleration is,

$t_1=\frac{\Delta y}{\left(\frac{v_0+v}{2}\right)}$

Here,

$t_1$ is duration at which the rocket has upward acceleration

Substitute $150\text{m}$ for $\Delta y$, $50.0\text{m/s}$ for $v_0$ and $55.7\text{m/s}$ for $v$ to find $t_1$.

$\begin{array}{c}{t}_1=\frac{150\text{m}}{\left(\frac{\left(50.0\text{m/s}\right)+\left(55.7\text{m/s}\right)}{2}\right)}\\ =2.84\text{s}\end{array}$

Thus, the time interval at which the rocket has upward acceleration is $2.84\text{s}$.

The formula used to calculate the interval at which the rocket moves upwards after the engine stops is,

$t_2=\frac{\Delta y\text{'}}{\left(\frac{v+v_f}{2}\right)}$

Here,

$t_2$ is duration at which the rocket moves upwards after the engine stops

Substitute $158\text{m}$ for $\Delta y\text{'}$, $55.7\text{m/s}$ for $v$ and $0$ for $v_f$ to find $t_2$.

$\begin{array}{c}{t}_2=\frac{158\text{m}}{\left(\frac{\left(55.7\text{m/s}\right)+\left(0\right)}{2}\right)}\\ =5.67\text{s}\end{array}$

Thus, the time interval at which the rocket moves upward after the engine stops is $5.67\text{s}$.

The formula used to calculate the total time of the upward flight is,

$t=t_1+t_2$

Here,

$t$ is total time of upward flight

Substitute $2.84\text{s}$ for $t_1$ and $5.67\text{s}$ for $t_2$ to find $t$.

$\begin{array}{c}t=2.84\text{s}+5.67\text{s}\\ =8.51\text{s}\end{array}$

Thus, the time taken by the rocket to reach the maximum height $8.51\text{s}$.

Conclusion:

The time taken by the rocket to reach the maximum height $8.51\text{s}$.

To determine

The total time the rocket stays in air.

Answer to Problem 53P

The total time the rocket stays in air $16.4\text{s}$.

Explanation of Solution

The formula used to calculate the total time of the downward flight is,

$t_3=\sqrt{\frac{2\left(-h_{\max }\right)}{a\text{'}}}$

Here,

$t_3$ is the total time of the downward flight is,

Substitute $308\text{m}$ for $h_{\max }$ and $-9.80{\text{m/s}}^2$ for $a\text{'}$ to find $t_3$.

$\begin{array}{c}{t}_3=\sqrt{\frac{2\left(-308\text{m}\right)}{\left(-9.80{\text{m/s}}^2\right)}}\\ =7.93\text{s}\end{array}$

Thus, the total time of the downward flight is $7.93\text{s}$.

The formula used to calculate the total time of flight is,

$T=t+t_3$

Here,

$T$ is the total time of flight of the rocket

Substitute $8.51\text{s}$ for $t$ and $7.93\text{s}$ for $t_3$ to find $T$.

$\begin{array}{c}T=8.51\text{s}+7.93\text{s}\\ =16.4\text{s}\end{array}$

Thus, the total time the rocket stays in air $16.4\text{s}$.

Conclusion:

The total time the rocket stays in air $16.4\text{s}$.