Question

Given a projectile's maximum horizontal range on level ground is R = Vo^2sin 20/g, Keeping

Answer 1

Answer:

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°