Final answer:
To find the launch angle where the horizontal range is equal to three times the maximum height, we apply projectile motion equations R = (v^2 × sin(2θ)) / g and H = (v^2 × sin^2(θ)) / (2g). By setting R = 3H, simplifying, and using trigonometry, we get tan(θ) = 3, leading to an angle of 71.57 degrees. Considering the complementary angles, the closest matching answer is 75 degrees.
Step-by-step explanation:
The question asks at what angle a projectile should be fired such that its horizontal range is equal to three times its maximum height. To solve this, we can use the fact that the range R of a projectile launched at an angle θ with initial velocity v is given by the equation R = (v^2 × sin(2θ)) / g, where g is the acceleration due to gravity. The maximum height H can be found using H = (v^2 × sin^2(θ)) / (2g).
According to the given conditions, we can write the relation R = 3H, leading to the equation (v^2 × sin(2θ)) / g = 3 × (v^2 × sin^2(θ)) / (2g). After simplification and applying trigonometric identities, we obtain tan(θ) = 3, which gives an angle θ = arctan(3). The value of arctan(3) is approximately 71.57 degrees. However, since this is not one of the options given and we know that complementary angles yield the same range, the correct answer is 90 degrees - 71.57 degrees = 18.43 degrees. Complementary to this angle is 71.57 degrees, and so the closest option given in the question is 75 degrees.