Final answer:
The angle of projection needed for a projectile to have its maximum height equal to its horizontal range is 45 degrees. At this angle, the projectile's initial speed is equally distributed between its vertical and horizontal components, optimizing both height and range.
Step-by-step explanation:
The question involves determining the angle of projection for a projectile with a certain initial speed such that its maximum height equals its horizontal range. To achieve a condition where the maximum height is equal to the horizontal range, the angle of projection must be 45°. For any other angle, either the maximum height or the range will be greater. At the 45° projection angle, both the vertical and horizontal components of the initial velocity are equal, which, owing to the symmetry of projectile motion under uniform gravity, leads to equal ranges and heights.
The formula for the range of a projectile is R = (v^2 * sin(2θ)) / g, where R is the range, v is the initial speed, θ is the launch angle, and g is the acceleration due to gravity. To find the height, we can use the following formula: H = (v^2 * sin^2(θ)) / (2g). Setting the angle to 45° ensures that sin(2θ) = sin(90°) = 1, and sin^2(θ) also reaches its maximum value of 1 / 2. For maximum range and height to be the same, sine terms must be equal, which only occurs at 45°.