Respuesta :
Answer:
1) θ = 45, b) R₄₀ = R₅₀ = (v₀² / g) 0.985, R₂₅ = R₆₅ = (v₀² / g) 0.766
3) θ = 90º, d) y₂₅ = (v₀ / 2g) 0.179, y₄₅ = (v₀ / 2g) 0.5, y₆₅ = (v₀ / 2g) 0.821
Explanation:
This is a projectile throwing exercise, let's use the relations
R = [tex]\frac{v_o^2 \ sin \ 2\theta }{g}[/tex]
1) the angle for a maximum reach (R) is θ = 45
R = v₀²/g
2) let's compare the scopes
θ = 25º
R₂₅ = (v₀² /g) sin 2 25
R₂₅ = (v₀² /g) 0.766
θ = 65º
R₆₅ = (v₀² / g) sin 2 65
R₆₅ = (v₀² / g) 0.766
θ = 40º
R₄₀ = (v₀²/g) sin 2 40
R₄₀ = (v₀²/ g) 0.985
θ = 50
R₅₀ = (v₀²/g) sin 2 50
R₅₀ = (v₀² / g) 0.985
we can see that it has the same scope
R₄₀ = R₅₀ = (v₀² / g) 0.985
R₂₅ = R₆₅ = (v₀² / g) 0.766
3) the maximum height is reached when the vertical speed is zero
[tex]v_y^2 = v_{oy}^2 - 2 g y[/tex]
v_y = 0
y = [tex]\frac{v_{oy}^2}{2g}[/tex]
y = (v₀² /2g) ( sin θ)²
therefore the height is maximum for θ = 90º
4) θ = 25º
y₂₅ = (v₀ / 2g) (sin 25)²
y₂₅ = (v₀ / 2g) 0.179
θ = 45º
y₄₅ = (v₀ / 2g) (sin 45)²
y₄₅ = (v₀ / 2g) 0.5
θ = 65º
y₆₀ = (v₀ / 2g) (sin 65)²
y₆₅ = (v₀ / 2g) 0.821
5) It is appreciated that as the launch angle increases the height increases, the maximum is when the movement is vertical
For the launch angle, the reach has a maximum for the angle of 45º, this is due to the symmetry of the sine function around this value.
