A model rocket is launched straight up into the air from ground level. The rocket engine burns for 3.2 s and provides a net upward acceleration during that time period of 27 m/s^{2}. How high does the rocket rise before it starts down again?
In this problem, you are asked to describe the motion (how high) of the rocket. Whenever you are asked to describe the motion of an object without worrying about the cause of that motion, you have a kinematics problem.
There are three key kinematic equations. If you carefully select the equation which most directly describes the situation in your problem, you will not only solve the problem in fewer steps but also understand it better. The three equations, written for motion in the y-direction, are:
1. y = y_{0} + v_{0}Δt + ½ a(Δt)^{2} (relates position and time)
2. v = v_{0} + aΔt (relates velocity and time)
3. v^{2} = v_{0}^{2} + 2a(Δy) (relates velocity and position)
These equations only apply for periods of motion in which acceleration is constant! In this problem, the acceleration of the rocket changes from one constant value to another at Point 2. Therefore, you will have to work the problem in two parts.
a) To find the distance the rocket rises in the first 3.2 seconds (from Point 1 to Point 2), relate position and time: equation 1 with a = 27 m/s^{2}.
b) To find the distance the rocket rises between Point 2 and its maximum height (y-velocity = 0), relate position and velocity: equation 3 with a = -9.8 m/s^{2}. (You might also recognize that to use equation 3 you will need to know the velocity at Point 2, or the velocity at 3.2 seconds. This requires equation 2 between Points 2 and 3. If you don’t catch that at this stage, that is fine. You will recognize it as you work the problem.)
Step 1:
You now know how high the rocket flew during the period of time while the engine was burning. Scroll down to find how much further the rocket went before coming to a stop at the top of the motion.
------------------------------------------------------------------------
Step 2:
Now that you know how high the rocket flies while the engine burns, you need to find how much further it goes to reach its maximum height: in other words, the distance between Point 2 and Point 3. All equations that relate Points 2 and 3 require you to know the velocity of the rocket at Point 2, so this must be done as an intermediate step.
You now have the information needed to solve kinematic equations for Point 3. Scroll down to continue with the solution.
------------------------------------------------------------------------
Step 3:
The rocket travels upward for 140 m between Points 1 and 2, and an additional 380 m between Points 2 and 3, so it travels a total of (140 m + 380 m) = 520 m to its highest point.
You do not have direct information about the amount of energy that the rocket engine converts from stored chemical energy (in the engine) to kinetic energy (of the rocket.) Therefore, it is most direct to approach this question with kinematics.
The rocket was launched from the ground and I want to know the total distance it rises, so I chose the ground to be the y = 0 point.
This is just for convenience in emphasizing that Point 3 is the top of the motion. Point 3 really is directly above point 1.
The problem states that there is a net upward acceleration while the rocket engine burns. The rocket is moving upward when the engine runs out of fuel, and so it will then begin to slow because gravity is now the only force acting on the rocket. In other words, for the first period of the problem (Point 1 to Point 2) the rocket experiences an acceleration of 27 m/s^{2}. After that (Point 2 to Point 3) it experiences an acceleration of -9.8 m/s^{2}. I marked Point 2 on the figure because acceleration changes there, even though I don’t know anything about the rocket’s position or velocity at that point.
The t in the kinematic equations refers to the time interval between the two points in the equation, with y_{0} occurring at the earlier time. I use Δt rather than t to be explicit that this is a time interval (t – t_{0}) and not a point in time.
Some text books will give more than three kinematic equations - for example, they may provide “range equations” or different versions of the equations for horizontal and vertical motion. Those are just these three equations solved for special cases. In my view, it is better to know and understand three equations that will always work than to memorize many equations which are specific to certain situations only.
You need to find how high the rocket travels while the engine is burning (while acceleration is 27 m/s^{2}.) This is the period spanned by Points 1 and 2, or the first 3.2 s of the motion.
The problem states that the net acceleration of the rocket while the engine burns is 27 m/s^{2}. Net means that this is the resulting acceleration from all forces on the rocket, so even though we know that gravity acts as the rocket flies up its effects are already taken into account in the stated value of 27 m/s^{2}. Even if the word “net” was not present, the acceleration that we observe for an object is the resulting acceleration from all forces acting. So for the 3.2 s between Points 1 and 2, a = 27 m/s^{2}.
Only two significant figures were given in the text of the problem, so only two significant figures are included in the solution.
In order to learn about the motion between Points 2 and 3, you first need to know the velocity of the rocket at Point 2. The rocket reaches its velocity because of the acceleration it experiences for 3.2 seconds between Points 1 and 2. Therefore, the appropriate points to consider are 1 and 2.
The problem states that the net acceleration of the rocket while the engine burns is 27 m/s^{2}. Net means that this is the resulting acceleration from all forces on the rocket, so even though we know that gravity acts as the rocket flies up its effects are already taken into account in the stated value of 27 m/s^{2}. Even if the word “net” was not present, the acceleration that we observe for an object is the resulting acceleration from all forces acting. So for the 3.2 s between Points 1 and 2, a = 27 m/s^{2}.
You need to find how high the rocket rises after the engine turns off, or the difference in y-position between Point 2 and Point 3. Therefore, “y_{0}” and “y” are “y_{2}” and “y_{3}” respectively. At Point 3, the rocket has come to a stop so v_{3} = 0. The speed of the rocket at Point 2 (86 m/s) was found in the previous step. If you didn’t recognize the need for that step when you started the problem, you would recognize it now.
The acceleration of the ball (due to gravity) is downward, which is the negative y-direction as chosen in this problem. Because a, v, and y are all vector quantities, their values are given a + or – sign to represent direction whenever they are filled into any of the kinematic equations.
Only two significant figures were given in the text of the problem, so only two significant figures are included in the solution.
You do not have direct information about the amount of energy that the rocket engine converts from stored chemical energy (in the engine) to kinetic energy (of the rocket.) Therefore, it is most direct to approach this question with kinematics.
The rocket was launched from the ground and I want to know the total distance it rises, so I chose the ground to be the y = 0 point.
The first point described in the problem is when the rocket leaves the ground, which was chosen to be y = 0. The rocket is at rest until launch and so velocity at this point is also zero.
The problem states that there is a net upward acceleration while the rocket engine burns. The rocket is moving upward when the engine runs out of fuel, and so it will then begin to slow because gravity is now the only force acting on the rocket. In other words, for the first period of the problem (Point 1 to Point 2) the rocket experiences an acceleration of 27 m/s^{2}. After that (Point 2 to Point 3) it experiences an acceleration of -9.8 m/s^{2}. I marked Point 2 on the figure because acceleration changes there, even though I don’t know anything about the rocket’s position or velocity at that point.
We know that the rocket continues to move upward after the engine has stopped burning, until gravity has slowed it to a stop. At that point, velocity in the y-direction is zero and the rocket will begin to speed up as it falls back down again.
In this case, the motion of the rocket is strictly vertical. You only have y-direction motion. Therefore, you do not need to divide any of the variables into their x- and y-components.
I drew the motion of the ball as a parabola, rather than straight up and down, only so that I could clearly mark points 1, 2, and 3 on the drawing and make clear that Point 2 is the top of the motion.
You need to find how high the rocket travels while the engine is burning (while acceleration is 27 m/s^{2}.) This is the period spanned by Points 1 and 2, or the first 3.2 s of the motion.
The problem states that the net acceleration of the rocket while the engine burns is 27 m/s^{2}. Net means that this is the resulting acceleration from all forces on the rocket, so even though we know that gravity acts as the rocket flies up its effects are already taken into account in the stated value of 27 m/s^{2}. Even if the word “net” was not present, the acceleration that we observe for an object is the resulting acceleration from all forces acting. So for the 3.2 s between Points 1 and 2, a = 27 m/s^{2}.
v_{1}Δt = (0)(3.2 s) = 0
Only two significant figures were given in the text of the problem, so only two significant figures are included in the solution.
In order to learn about the motion between Points 2 and 3, you first need to know the velocity of the rocket at Point 2. The rocket reaches its velocity because of the acceleration it experiences for 3.2 seconds between Points 1 and 2. Therefore, the appropriate points to consider are 1 and 2.
You need to find how high the rocket rises after the engine turns off, or the difference in y-position between Point 2 and Point 3. Therefore, “y_{0}” and “y” are “y_{2}” and “y_{3}” respectively. At Point 3, the rocket has come to a stop so v_{3} = 0. The speed of the rocket at Point 2 (86 m/s) was found in the previous step. If you didn’t recognize the need for that step when you started the problem, you would recognize it now.
Only two significant figures were given in the text of the problem, so only two significant figures are included in the solution.
The acceleration of the ball (due to gravity) is downward, which is the negative y-direction as chosen in this problem. Because a, v, and y are all vector quantities, their values are given a + or – sign to represent direction whenever they are filled into any of the kinematic equations.
No. While the rocket engine burns, it provides a force greater than gravity and so the rocket has a net upward acceleration. However, when the rocket engine runs out of fuel, the rocket is traveling upward. It will continue to travel up, slowing in speed as it goes, until gravity (providing a downward acceleration of 9.8 m/s^{2}) brings it to a stop.
It is correct that while the engine is burning both the engine and the gravity put a force on the rocket. However, the problem stated that the net acceleration of the rocket is 27 m/s^{2}. Net means that this is the resulting acceleration from all forces on the rocket, so the effects of gravity are already taken into account in the stated value of 27 m/s^{2}. Even if the word “net” was not present, the acceleration that we observe for an object is the resulting acceleration from all forces acting.
You are right. Within the first half second or so the rocket engine provides a very large upward acceleration which then drops to a smaller value that remains roughly constant until the engine runs out of fuel. I chose an average value of acceleration which gives the same maximum height for a rocket in order to simplify the math required in this problem.
All that matters for describing motion is that you explicitly state where your reference point is. In this case, you know that the rocket starts at rest from ground level and you are asked how high it goes above the ground. So picking y = 0 at ground level is a natural choice.
This is just for convenience in emphasizing that Point 3 is the top of the motion. Point 3 really is directly above point 1.
The kinematic equations relate position and velocity at any two points in the motion. Because many problems provide or ask about information at more than two points, I identify each point with its own subscript to keep them straight. I can then pick any two points to be y_{0} and y when I use the equations. The t in the kinematic equations refers to the time interval between the two points in the equation, with y_{0} occurring at the earlier time. I use Δt rather than t to be explicit that this is a time interval (t – t_{0}) and not a point in time
The t in the kinematic equations refers to the time interval between the two points in the equation, with y_{0} occurring at the earlier time. I use Δt rather than t to be explicit that this is a time interval (t – t_{0}) and not a point in time.
Some text books will give more than three kinematic equations—for example, they may provide “range equations” or different versions of the equations for horizontal and vertical motion. Those are just these three equations solved for special cases. In my view, it is better to know and understand three equations that will always work than to memorize many equations which are specific to certain situations only.
The rocket in this problem experienced a change in acceleration when the engine stopped burning. The kinematic equations are only valid for periods of constant acceleration. (They were derived from the definitions of velocity and acceleration plus the condition of constant acceleration.) Therefore, you had to work the problem in two parts.
During the first part of the problem, the observed acceleration of the rocket was 27 m/s^{2} and the rocket accelerated from zero to 86 m/s while traveling a distance of 140 m. Knowing these values at Point 2 allowed you to then learn about Point 3, with the understanding that the observed acceleration of the rocket between Points 2 and 3 was -9.8 m/s^{2}.
Because the initial acceleration is almost a factor of 3 greater than the final acceleration, it makes sense that the distance covered in the second part of the motion is almost a factor of three greater than in the initial leg. In both cases, the rocket went between speeds of 0 and 86 m/s, but it took a much longer time to do so after the engine shut off.