I'm doing a piece of Physics work from which the experiment is (How the diameter of a cone affects the time it takes for it to reach the ground).
There's marks for going above and beyond the curriculum but i'm stuck, i've related it with distance/time = speed but not for the ones below
Are there any suggestions on how I could relate the Drag equation or the Aerodynamic equation(s) to this. 5 stars for the best answer, instantly. Fairly indepth please.
There's marks for going above and beyond the curriculum but i'm stuck, i've related it with distance/time = speed but not for the ones below
Are there any suggestions on how I could relate the Drag equation or the Aerodynamic equation(s) to this. 5 stars for the best answer, instantly. Fairly indepth please.

Okay, here's what you want to say. First, just note that the cone will fall at an approximate acceleration of g (9.8 m/s^2) downward. Thus, the time (t) to fall a height (h) is given by t = Sqrt(2*h/g). Theoretically, the diameter would have no effect. That's all you need for full points.
Now for the above and beyond stuff. The drag force is given by the equation D = 1/2*p*V^2*S*Cd, where p is the density of air, V is the velocity, S is the area, and Cd is the coefficient of drag. If dia is your diameter, then S = pi*(dia/2)^2. Then D = 1/2*p*v^2*pi*dia^2/4*Cd. You can basically reduce this with a big constant (we'll call it k), to get D = k*dia^2*v^2.
Then your acceleration, instead of g, becomes a = (mg  k*dia^2*v^2)/m = g  k/m*dia^2*v^2. Unfortunately, now things get complicated because you have a second order differential equation. Personally, I would stop at this point, and say that since the decrease in acceleration is related to the square of the diameter, that cones with larger diameters will take longer to reach the ground.
***Extra stuff***
If you're weird, and you really want to go "above and beyond", you could say the mass of the cone increases with the volume (V=1/12*pi*dia^2*h), which increases with the diameter. You could also cite this website http://www.aerospaceweb.org/question/aer… for the equation of a cone's coefficient of drag. Cd = .0112*e + .162, where e is the half vertex angle in degrees, which will also increase with the diameter. You could then write everything into a complex equation solver, like Wolfram Mathematica, and plot the function. But don't do this unless you hate yourself.
Now for the above and beyond stuff. The drag force is given by the equation D = 1/2*p*V^2*S*Cd, where p is the density of air, V is the velocity, S is the area, and Cd is the coefficient of drag. If dia is your diameter, then S = pi*(dia/2)^2. Then D = 1/2*p*v^2*pi*dia^2/4*Cd. You can basically reduce this with a big constant (we'll call it k), to get D = k*dia^2*v^2.
Then your acceleration, instead of g, becomes a = (mg  k*dia^2*v^2)/m = g  k/m*dia^2*v^2. Unfortunately, now things get complicated because you have a second order differential equation. Personally, I would stop at this point, and say that since the decrease in acceleration is related to the square of the diameter, that cones with larger diameters will take longer to reach the ground.
***Extra stuff***
If you're weird, and you really want to go "above and beyond", you could say the mass of the cone increases with the volume (V=1/12*pi*dia^2*h), which increases with the diameter. You could also cite this website http://www.aerospaceweb.org/question/aer… for the equation of a cone's coefficient of drag. Cd = .0112*e + .162, where e is the half vertex angle in degrees, which will also increase with the diameter. You could then write everything into a complex equation solver, like Wolfram Mathematica, and plot the function. But don't do this unless you hate yourself.