Answer:
15 degrees
Step-by-step explanation:
Let's assume that the initial velocity of the projectile is v0, the angle of projection is θ, and g is the acceleration due to gravity.
The maximum height reached by the projectile can be calculated using the formula:
H = (v0² sin²θ)/(2g)
The range of the projectile can be calculated using the formula:
R = (v0² sin2θ)/g
We want to find the angle θ at which the range is half the maximum height, i.e., R = H/2. Substituting the equations for H and R, we get:
(v0² sin²θ)/(2g) = (v0² sin2θ)/2g * (1/2)
sinθ cosθ = 1/4
Now, we can use the identity sin2θ = 2 sinθ cosθ to write:
sin2θ = 2 sinθ cosθ = 2 * (1/4) = 1/2
Taking the inverse sine of both sides, we get:
2θ = 30°
θ = 15°
Therefore, the angle at which the range of the projectile is equal to half the maximum height is 15 degrees.
I hope this helped :)