Answer:
v₀ = 11.72 m / s
Explanation:
This is a missile launch exercise. Let's write the relations for the final position
x = v₀ₓ t
y = [tex]v_{oy}[/tex] t - ½ g t²
Let's use trigonometry to find the initial velocities
sin 45 = v_{oy} / v₀
cos 45 = v₀ₓ / v₀
v_{oy} = v₀ sin 45
v₀ˣ = v₀ cos45
we substitute
x = v₀ cos 45 t
y = v₀ sin 45 t - ½ g t²
in this equation system we have two unknowns and two equations for which it can be solved, let's substitute the values
7 = v₀ 0.707 t
3.5 = v₀ 0.707 t - ½ 9.8 t²
we clear the time in the first equation and substitute in the second
3.5 = v₀ 0.707 (7 / v₀ 0.707) - 4.9 (7 / v₀ 0.707) 2
3.5 = 7 - 480.345 / v₀²
480.345 / v₀² = 7-3.5
v₀² = 480.345 / 3.5
v₀ = √ 137.24
v₀ = 11.72 m / s