Take the x-axis as horizontal and the y-axis as vertical. A gun at the origin can fire at any angle into the first quadrant (x, y ≥ 0) with a fixed muzzle velocity v. Assuming the only force on the pellet after firing is gravity (acceleration g), which points in the first quadrant can the gun hit?

**Solution**

Let the angle of the gun to the x-axis be θ. Then the equations of motion are: x = t v cos θ, y = t v sin θ - 1/2 g t^{2}. So the pellet moves along the parabola y = x tan θ - x^{2} g/2v^{2} sec^{2}θ (*).

We can view (*) as an equation for θ given x, y. Put k = tan θ, then the equation becomes g/2v^{2} x^{2} k^{2} - x k + y + gx^{2}/2v^{2}. This has real roots iff x^{2} ≥ 2g/v^{2} x^{2}y - g^{2}/v^{4} x^{4} and hence iff y ≤ 2g^{2}/v^{2} - x^{2}/v^{4} x^{2} (**). Since x ≥ 0, we see directly from the quadratic that the sum and the product of the roots are both non-negative, so (**) is the condition for the equation to have at least one non-negative root in k and hence at least one root for θ in the range 0 to π/2. Thus the gun can hit points in the first quadrant under (or on) the parabola given by (**).

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002