The horizontal range of an oblique projectile is equal to the distance through which a projectile has to fall freely from rest to acquire a velocity equal to the velocity of projection in magnitude. The angle of projection isA75oB60oC45oD30o

The horizontal range of a projectile is given by the formula R = (v^2/g) * sin(2θ), where v is the initial velocity, g is the acceleration due to gravity, and θ is the angle of projection.

The velocity acquired by a freely falling object is given by the formula v = sqrt(2gh), where h is the height from which the object falls.

Given that the horizontal range is equal to the distance through which a projectile has to fall freely from rest to acquire a velocity equal to the velocity of projection in magnitude, we can equate the two formulas:

v^2/g = 2gh v^2 = 2ghg v = sqrt(2ghg)

Substituting this back into the range formula gives:

R = (2ghg/g) * sin(2θ) R = 2h * sin(2θ)

Since the range is equal to the height, we can equate the two:

2h = 2h * sin(2θ) 1 = sin(2θ)

The angle whose sine is 1 is 90 degrees, so 2θ = 90 degrees, and θ = 45 degrees.

Therefore, the angle of projection is 45 degrees, so the answer is C.