Given a projectile's maximum horizontal range on level ground is r = vo^2sin 20/g, keeping
the initial velocity vo constant, at what launch angle or angles will the projectile land a
distance r/2 ? (2 points)

15°, 75°

Step-by-step explanation:

The range of a projectile on level ground is:

R = v₀² sin(2θ) / g

For a range of R/2 at the same initial velocity:

R/2 = v₀² sin(2A) / g

Dividing the second equation by the first:

(R/2) / R = sin(2A) / sin(2θ)

1/2 = sin(2A) / sin(2θ)

sin(2A) = 1/2 sin(2θ)

The maximum range is when θ = 45°.

sin(2A) = 1/2 sin(90°)

sin(2A) = 1/2

2A = 30°, 150°

A = 15°, 75°

proofs involving isosceles triangles often require special consideration because an isosceles triangle has several distinct properties that do not apply to normal triangles.(more about triangle types) therefore, when you are trying to prove that two triangles are congruent, and one or both triangles, are isosceles you have a few theorems that you can use to make your life easier.

step-by-step explanation:

an isosceles triangle has two congruent sides and two congruent angles. the congruent angles are called the base angles and the other angle is known as the vertex angle. ∠bac and ∠bca are the base angles of the triangle picture on the left. the vertex angle is ∠abc