A boy stands at the peak of a hill which slopes downward uniformly at angle . At

Question

A boy stands at the peak of a hill which slopes downward uniformly at angle . At what angle from the horizontal should he throw a rock so that it has the greatest range?

Explanation / Answer

If we choose the origin of our coordinate system to be at the launch point, the equation of the slope is yslope = - x tan f (1) The x coordinate of the rock as a function of time is xrock = vxt = v0xt = v0t cos q (2) where vx = v0x = v0 cos q is the x velocity of the rock, which is constant. The y coordinate of the rock as a function of time is yrock = vy0t - (1/2)gt2 = v0t sin q - (1/2)gt2 (3) where vy0 = v0 sin q is the initial y velocity of the rock. From (2), we see that t = xrock / v0 cos q Substituting into (3), we get yrock = xrock tan q - (1/2)g(xrock / v0 cos q)2 or simply yrock = x tan q - (1/2)g(x / v0 cos q)2 (4) The point at which the rock hits the slope is given by equating yslope and yrock, and solving for x. Using (1) and (4), yslope = yrock => - x tan f = x tan q - (1/2)g(x / v0 cos q)2 Dividing both sides by x, - tan f = tan q - (1/2)g(1 / v0 cos q)2x Solving for x, x = (2/g)(v02 cos2 q)(tan q + tan f) = (2v02/g)(cos2 q)(tan q + tan f) To maximize x, we take the derivative of x with respect to q, set it equal to zero, and solve for x. dx/dq = - (2v02/g)(2 cos q sin q)(tan q + tan f) + (2v02/g)(cos2 q)(sec2 q) = 0 => - (2 cos q sin q)(tan q + tan f) + 1 = 0 => - 2 sin2 q - 2 cos q sin q tan f + 1 = 0 (5) We now make use of the following trigonometric identities: sin 2q = 2 sin q cos q (6) cos 2q = cos2 q - sin2 q (7) 1 = cos2 q + sin2 q (8) Subtracting (7) from (8), we get 1 - cos 2q = 2 sin2 q => sin2 q = (1 - cos 2q) / 2 (9) Using (6) and (9), we can rewrite (5) as - 1 + cos 2q - sin 2q tan f + 1 = 0 => cos 2q - sin 2q tan f = 0 => 1 - tan 2q tan f = 0 => q = (1/2) tan-1(1 / tan f)

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