Answer: 75.96... degrees from the horizontal. Let the projectile be launched with speed v at an angle θ degrees above the horizontal. Then its vertical speed component is v sin θ, and its horizontal component is v cos θ. The time of flight is t = 2 v sin θ / g, so that the range R is given by R = v t cos θ = 2 v^2 sin θ cos θ / g. The maximum height is given by H = (v sin θ)^2 / (2 g). (You may think of this is as merely an application of the standard result that, dropping from rest in the second half of the flight, the downward speed u = v sin θ must be related to the downward acceleration a and distance d travelled by u^2 = 2ad. In this context, where a = g and d = H, that is equivalent to the conservation of kinetic plus potential energy, of course.) If R = H, then v^2 sin^2 θ / (2 g) = 2 v^2 sin θ cos θ / g. Therefore sin θ / cos θ = tan θ = 4. Thus θ = arctan (4) = 75.96... degrees.
Posted on: Sun, 07 Jul 2013 06:43:45 +0000
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