Rocket Project:
Cover Letter:
During this project we were tasked to design and build a rocket that was able to launch and deploy a parachute, then calculate max height, time of max height, initial velocity, and theoretical flight time using the trig function, SOH CAH TOA. Factors we needed to consider are:
In this experiment we had to use quadratic functions to help us find projectile motion. A quadratic function is any function where x to the power of 2 is the highest power. This function always makes a parabola which is the same shape as when you throw an object through the air. We had to first make a blueprint of the rocket that we wanted to build. Then we started on the build and tested our rocket and refined as much as possible to get the best outcome.
- Position
- Position being a place where an object has been placed.
- Velocity
- Velocity being the rate of which an object changes position
- Acceleration
- Acceleration being the increase of speed of an object.
In this experiment we had to use quadratic functions to help us find projectile motion. A quadratic function is any function where x to the power of 2 is the highest power. This function always makes a parabola which is the same shape as when you throw an object through the air. We had to first make a blueprint of the rocket that we wanted to build. Then we started on the build and tested our rocket and refined as much as possible to get the best outcome.
Calculations:
Max Height:
To find max height first I had to find the measurements that I already knew. I knew that the height of the inclinometer was 4.75 ft off of the ground and it was positioned 200 ft away from the rocket launcher. I also knew that the inclinometer was at a 26 degree angle at the point of maximum height. From there I drew out a triangle and labeled it (the angle at 26 degrees and the side adjacent from the angle at 200 ft). Now all I needed to do was to use SOH-CAH-TOA to find the max height. Since I knew the angle and the adjacent side, I knew that I had to use TOA (Tangent of (x) = the opposite over the hypotenuse). Then I plugged in the numbers to the equation
Tan(26) = x/200
To get x by itself I had to multiply by 200 (200 tan(26)=x)
After we got that we just plugged it into Desmos and got 101.75 as our maximum height.
Time of max height:
To find the time of max height of our rocket we found the video of our rocket when we launched it at the exhibition. We knew that a phone shoots at 30 fps. So we counted the frames from the initial launch to the max height. We counted 43 frames to the max height. Then we divided 42 by 30 to get us the time of mac height = 1.4 seconds. After that we had to find the total flight time. So we counted the frames from the initial launch to when the rocket lands on the ground, we counted 126 frames then we divided by 30 to get the total fight time = 4.2 seconds.
Initial velocity:
To solve for the initial velocity we took all of our constants and variables that we already knew to create a quadratic function. We used the max height (which was 101.75 ft) our time of max height (1.4 seconds) to solve for our initial velocity. Then we plugged all of our constants and variables into the equation 101.75=-1/2(32)(1.4)+Vo(1.4)+1.5 we had got Vo by itself. Our first step was to simplify the equation. First we multiplied -½ by 32 to get -16. Next we had to multiply -16 by (1.4)2 to get -31.36. We then subtracted 1.5 from each side to get -29.86. Our next step was to get Vo by itself so we added 29.86 to each side, that got us 131.61. We finally divided by 1.4 on each side to get our answer for Vo= 94.
Theoretical flight time:
We created a standard form of a quadratic ( h(t)=½ (32)(t)^2 + 94t+1.5 ) to find A(-16), B(94), and C(1.5) we then plugged those numbers into the quadratic formula:
We plugged our numbers into a quadratic calculator to get our theoretical flight time which was 5.89 seconds
To find max height first I had to find the measurements that I already knew. I knew that the height of the inclinometer was 4.75 ft off of the ground and it was positioned 200 ft away from the rocket launcher. I also knew that the inclinometer was at a 26 degree angle at the point of maximum height. From there I drew out a triangle and labeled it (the angle at 26 degrees and the side adjacent from the angle at 200 ft). Now all I needed to do was to use SOH-CAH-TOA to find the max height. Since I knew the angle and the adjacent side, I knew that I had to use TOA (Tangent of (x) = the opposite over the hypotenuse). Then I plugged in the numbers to the equation
Tan(26) = x/200
To get x by itself I had to multiply by 200 (200 tan(26)=x)
After we got that we just plugged it into Desmos and got 101.75 as our maximum height.
Time of max height:
To find the time of max height of our rocket we found the video of our rocket when we launched it at the exhibition. We knew that a phone shoots at 30 fps. So we counted the frames from the initial launch to the max height. We counted 43 frames to the max height. Then we divided 42 by 30 to get us the time of mac height = 1.4 seconds. After that we had to find the total flight time. So we counted the frames from the initial launch to when the rocket lands on the ground, we counted 126 frames then we divided by 30 to get the total fight time = 4.2 seconds.
Initial velocity:
To solve for the initial velocity we took all of our constants and variables that we already knew to create a quadratic function. We used the max height (which was 101.75 ft) our time of max height (1.4 seconds) to solve for our initial velocity. Then we plugged all of our constants and variables into the equation 101.75=-1/2(32)(1.4)+Vo(1.4)+1.5 we had got Vo by itself. Our first step was to simplify the equation. First we multiplied -½ by 32 to get -16. Next we had to multiply -16 by (1.4)2 to get -31.36. We then subtracted 1.5 from each side to get -29.86. Our next step was to get Vo by itself so we added 29.86 to each side, that got us 131.61. We finally divided by 1.4 on each side to get our answer for Vo= 94.
Theoretical flight time:
We created a standard form of a quadratic ( h(t)=½ (32)(t)^2 + 94t+1.5 ) to find A(-16), B(94), and C(1.5) we then plugged those numbers into the quadratic formula:
We plugged our numbers into a quadratic calculator to get our theoretical flight time which was 5.89 seconds
Reflection:
- What challenges did you encounter in this project and how did you navigate those challenges?
- What were your successes in this project and how would you relay those successes to next year’s sophomores?
- What was a turning point for you in this project and why?
- If you were to do this project again, what would you do differently and why?