During this unit, we explored the world of quadratic equations and completing squares. Our goal was to go as high as possible and to have our parachute deploy. We were disappointed with the height the rocket went. It was the same as our practice launch. The parachute was also wet due to over spill from the water jugs so it wouldn't deploy and possibly weighted down our rocket.Our rocket was launched off a stand about 1.5 ft high then pumped it full of pressurized water after that we launched and it accelerated 180 feet into the sky. We started off the rocket with smaller fins and a bottle cap nose cone. After a few tests we discovered that we need to make the fins bigger and lower them down the rocket more so it would increase drag at the bottom causing the rocket to go higher.
Max Height In order to calculate the time of max height, we would need an angle and the distance from the measurement of that angle to the rocket. We then need to plug these values into a right triangle with the distance from the anglenometer to the rocket being the adjacent. In our case, our adjacent would be 200, and our angle would be 52 degrees. We can then use SOH CAH TOA to calculate our max height, which on the triangle is our opposite. We can then do 200*tangent(52) = 255 which is our max height.
Time of Max Height In order to calculate time of max height you’ll need a video of the entire flight. You then want to count every frame up to when the rocket reaches its maximum height. An Iphone shoots at 30 frames per second, meaning that in order to find the time of max height in seconds we’ll need to divide our counted frames by 30. I counted 72 frames to my rockets max height and divided by 30 giving me a time of 2.4 seconds(72/30 = 2.4)
In order to calculate this, we’ll use the equation, H(t) = -½ (g)(t2) + V0 (t) + y0where h(t)= our max height (255 ft), V0 = Our initial velocity(?), t = our time of max height in seconds(2.4s), g = our gravity (32 ft/s2, and y0 = our starting height (1.5 ft). We can then plug these values into our equation and then from there start isolating our V0value which will look something like this: 255 = -½ (32)(2.4)2+ V0 (2.4) + 1.5 255 = (-16)(5.76) + V0 (2.4) +1.5 255 = -92.16 + V0 (2.4) + 1.5 +92.16 /2.4 -1.5 (255 + 92.16- 1.5) / 2.4 = V0 144.025 = V0
Theoretical Flight Time
In order to find our theoretical flight, we need to use our initial velocity formula and convert it into the quadratic formula. Our formula for initial velocity : H(t) = (-16)t2 +144t+1.5 in its standard quadratic form becomes (-16)(2.4)2+144 (2.4)+1.5. Converting it into the quadratic formula, we get x =-144 1442-4 (-16) (1.5)2(-16) which when solved for its’ 0’s becomes x=9.01 seconds which is 2 seconds off of our actual flight time of 11 seconds.