option C
45°
Equation to find the range of the motion:
R =[tex]\frac{u^2sin2\alpha }{2g}[/tex]
Largest value for sineθ is 1, with an input angle of 90 degrees, so we can see that for the longest-range, 2θ should be equal to 90 degrees.
Therefore, θ = 90°/2
θ = 45°
Thus, the largest horizontal distance covered is when an object is launched at an angle of 45 degrees.
