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A project for a physics course on projectile motion and numerical integration. It covers the calculation of trajectories for a canon ball with no air resistance using analytical and numerical methods. The document also includes the modification of the problem to include air resistance and the calculation of maximum altitudes for different initial velocities. Additionally, the document presents a similar problem for a rocket with air resistance and the calculation of maximum altitudes for different thrusts.
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Projectile Motion and Numerical Integration.
I.. Trajectory with no air resistance A. A canon ball is fired straight up with an initial velocity vo. Assuming no air resistance, write an expression for the position of the ball as a function of time y(t).
B. Now open the Excel file plotting the trajectory. Look at the different columns and see how they are calculated. For one pair of columns, y and v are calculated analytically using the equation from part a.
For the next pair, y and v are calculated numerically. This is done as follows.
First and initial position y(0) and velocity v(0) are defined. Knowing the net force on the ball, the acceleration may be calculated using Newton’s 2nd^ law. a(0) = F(0) / m
After a short interval of time dt, we can estimate the new velocity by assuming the acceleration was constant over that interval. v(1) = v(0) + a(0) * dt
Likewise, we can find the position at this time. y(1) = y(0) + v(0) * dt
This process is then repeated over and over again to find later values of a, v and y: a(n) = F(n) / m v(n+1) = v(n) + a(n) * dt y(n+1) = y(n) + v(n) * dt Currently, the data only runs to 16 seconds. To extend the calculation, just copy and past the last row. This is numerical integration. (Specifically, this particular method is called a Left Riemann Sum.) The process is illustrated below graphically. Note that as the interval dt decreases, the accuracy of the estimate improves.
a(t)
Area = ∆v dt
II. Canon ball trajectory with air resistance.
Now that we have a basic scheme for doing numerical integrals, it is straightforward to do other problems which are difficult or even impossible to solve analytically.
Modify this problem to include air resistance. The force of drag is given by (^1 2) ˆ d (^) 2 d F = − ρ v AC v
ρ = density of air or 1.292 kg/m^3 v = velocity A = cross sectional area of the projectile Cd = drag coefficient (0.1 for a smooth sphere)
A. Assume the mass of the ball is 10 kg, the initial velocity is 200 m/s, and the radius of the ball is 0.1 m. B. On the same graph, plot the trajectory of the ball with and without air resistance. What is the maximum altitude for each case? C. Increase the initial velocity to 400 m/s. What are the new maximum altitudes?
III. Rocket trajectory with air resistance.
Now consider a rocket with an initial mass of 5 kg fired from rest straight up. The rocket engine burns for 3 seconds and consumes 5 kg of fuel. (i.e., the rocket loses 5 kg of mass over 3 seconds.) The thrust of the rocket while it is burning is 100 N. Assume the drag coefficient is 0.25, and the radius of the rocket is 0.1 m.
A. Plot the trajectory of the rocket with and without air resistance. What are the maximum altitudes? B. Increase the thrust of the rocket to 400 N. Plot the new trajectories of the rocket. What are the new maximum altitudes?
v(t)
Area = ∆y
dt