Question

A projectile is fired in such a way that its horizontal range is equal to 13.5 times its maximum height. What is the angle of projection?

Answer 1

Answer:

Maximum height: h  = v0² sin² α / 2 g

Horizontal range: x = v0² sin 2 α / g

13.5 v0² sin² α /2 g  = v0² sin 2 α /g          / :  v0²/g

6.75 sin² α = 2 sin α cos α    / : sin α

6.75 sin α = 2 cos α    / : cos α

6.75 tan α = 2

tan α = 0.2963

α = tan^(-1) 0.2963

α = 16.504° ≈ 16.5°

The angle of projection is 16° 30`.

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Explanation:

Answer 2

Maximum height: h  = v0² sin² α / 2 g
Horizontal range: x = v0² sin 2 α / g
13.5 v0² sin² α /2 g  = v0² sin 2 α /g          / :  v0²/g
6.75 sin² α = 2 sin α cos α    / : sin α
6.75 sin α = 2 cos α    / : cos α
6.75 tan α = 2
tan α = 0.2963
α = tan^(-1) 0.2963
α = 16.504° ≈ 16.5°
The angle of projection is 16° 30`.