Table approach

We impose a coordinate system scaled in feet. We stand at the origin and the target is at (3,000, 0) . The hill extends from the origin back into the second quadrant. Then, we want to position the catapult at such a point that the path of the projectile will intersect the x-axis at (3000,0). Before moving the catapult, it makes sense to figure out where the projectile will land if we launch it from the origin.

We know this from Part 1, namely our maximum range when the catapult is at the origin will be 2,812.5 feet downrange. Thus our shot will fall about 87.5 feet short of the target. We present this analysis here again for completeness.

When fired form the origin, the projectile's position t seconds after being fired will be given by the parametric equations

x(t) = 300 (cos 45)t

y(t) = -16t^2 + 300 (sin 45) t

We are interested in the time(s) that make y(t)= 0 and the range at which this occurs. We know from Part 1 that the angle of elevation theta which give rise to the maximum range is Pi/4. Mathematica can then tell us the maximum range via the following code:

Again, from the above analysis we see that if our catapult is positioned at the origin, the projectile will hit 2,812.5 feet downrange. Thus our shot will fall about 187.5 feet short of the enemy camp which is out at (3000, 0).


The intuitive solution is to move the catapult closer to their camp. Since this horizontal translation does not affect the range of the catapult, if we fire it from (x, 0), it will land at (x + 2812.5, 0). However, the river prevents moving along the x-axis further than the point (30, 0). A projectile launched from this point will land at (2842.5, 0), still about 57.5 feet short of the target. We cannot reach our target by moving the catapult closer to the target. There seems to be no way to hit the target. And there isn't, by moving forward!

Now we examine the non-intuitive possibility--move the catapult backwards, up the hill. It is a very steep hill, and perhaps what we gain in range by elevating the gun will offset the horizontal loss of moving the gun away from the target. Because the hill is so steep, every horizontal foot we lose allows a vertical gain of square root of 3 feet. The following parametric equations describe the the projectile's position t seconds after being fired from the point (- a, b), where a > 0 and b > 0.

NOTE: We presume that our catapult still has an angle of elevation with the horizontal of Pi/4 for maximum range - although later we consider the possibility that another angle might give a further range.

We proceed to solve by creating a table of maximum ranges or landing points, using Mathematica to try several locations up the hill.

This table shows us that we can reach the target which is out at 3,000 ft. Our maximum landing point is around 3030 ft when k is somewhere between 700 and 900. So we can surely hit our target at 3,000 ft out, but exactly what value of k should we use? We now focus our table on the interval 400 to 500.

This table shows that moving the catapult back up the hill to the the point (-460, 460 Sqrt(3)) will shoot a projectile on the point (3000.12, 0). This means we have to move the catapult some 920 ft overland up the hill. If this hit is not direct enough, we can do better with closer examination of the interval -460 < x < - 450.