A projectile is launched on level ground. At the peak of its trajectory, it is 7.5 m high. It lands

10 m from where it was launched. At what angle was the projectile launched?

0.5099646386 radians

Explanation:

Δy = 7.5 m

Δx = 10 m

g = 9.8 m sec⁻²

From trigonometry,

Vy₀ = v₀ sin θ

Vy₀² = v₀² sin²θ

From the equations of constant acceleration kinematics,

Vy₀² = 2g Δy

Two things that are equal to the same thing are equal to each other, so

v₀² sin²θ = 2g Δy

v₀² sin²θ = 147 m² sec⁻²

The double angle formula for the sine,

sin²θ = 0.5 [1 - cos (2θ) ]

0.5 v₀² [1 - cos (2θ) ] = 147 m² sec⁻²

v₀² = 294 m² sec⁻² / [1 - cos (2θ) ]

The range equation,

Δx = v₀² sin (2θ) / (2g)

10 m = 294 m² sec⁻² sin (2θ) / { (2g) [1 - cos (2θ) ]}

1.5 sin (2θ) = 1 - cos (2θ)

1.5 sin (2θ) = 1 - √[1-sin² (2θ) ]

A convenient substitution.

u = sin (2θ)

2.25 u² = 1 - 2√[1-u²] + 1 - u²

3.25 u² - 2 = - 2√[1-u²]

1 - 1.625u² = √[1-u²]

1 - 3.25u² + 2.640625u⁴ = 1-u²

2.640625u⁴ - 2.25u² = 0

As long as u≠0,

2.640625u² - 2.25 = 0

2.640625u² = 2.25

u = 0.8520710059

θ = 0.5 arcsin u

θ = 0.5099646386 radians

Furthermore,

v₀ = 15.16666667 m/s