7th Putnam 1947 Problem A4 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?
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 (**). 7th Putnam 1947 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002