PHYS121 Winter 2000
A student wrote:

Could you help me begin problem 4-44? I'm not quite sure how to construct my answer.

Prof. Seidler responds:

OK, first be warned that this is a fairly complicated problem. The general framework for a solution will follow:
i. Derive an expression for the range R as a function of the elevation angle theta of the cannon.
ii. Find the angle giving the maximum range setting equal to zero the derivative with respect to theta of your result to i.

OK, so the next question, is how to get the necessary expression for R in terms of theta. I believe that one nice way to solve the problem is as follows:

1. choose a coordinate system whose origin is at the cannon, whose x-axis is along the surface of the hill, and whose y-axis is perpendicular to the surface of the hill.
2. Determine the acceleration due to gravity in this coordinate system {you should get ax=-g*sin(alpha) and ay=-g*cos(alpha) as the acceleration vector makes an angle of alpha with the negative y-axis.}
3. Determine the coordinates of the initial velocity vector vo in your chosen coordinate system.
4. You can now write down equation 4-12 in component notation (which means seperate equations for the x- and y-components of motion).
5. When the projectile hits the ground, y = 0. Use this together with your y-component equation in step 4 to determine the time at which the projectile will hit the ground. You should get t_hit=2*vo*sin(theta-alpha)/(g*cos(alpha))
6. Now substitute t_hit into your x-component equation from step 4 above. This is the desired expression for R in terms of theta.
7. Take a derivative of R with respect to theta, and set this equal to zero to find an extrema (maxima in this case). Solve this expression for theta. You should get theta = alpha + (1/2)*arccot(tan(alpha)). With a little trig thinking you will find that theta = alpha/2 + pi/4.