
Respuesta :
Answer:
answers the correct one is C
Explanation:
For this exercise we must use the projectile launch ratios
the expressions for the initial velocities are
      sin θ =v_{oy} / vo
      cos θ = v₀ₓ / vo
      v_{oy} = v₀ sin θ
      v₀ₓ = v₀ cos θ
Range and flight time are requested
the expression for the scope is
     R = [tex]\frac{ v_{o}^2 \ sin 2 \theta}{g}[/tex]
     Â
We calculate for each angle
θ = 45º
     R₄₅ = \frac{ v_{o}^2 \  sin  90}{g}
     R₄₅ = v₀² / g
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θ = 60º
     R₆₀ = \frac{ v_{o}^2 \  sin  120}{g}
     sin  120 = sin 60
     R₆₀60 = sin  60  R₄₅
as the sine function has values ​​between 0 and 1, the range for this angle is less
Flight time is twice the time it takes to reach maximum altitude
      v_y = v_{oy} - gt
at the point of maximum height there is no vertical velocity vy = 0
      t = v_{oy} / g
      t = v₀ sin θ / g
θ=45º angle
      t₄₅ = sin 45  v₀/g
      t₄₅ = 0.707 v₀/g
θ=60º angle
     t₆₀ = sin 60 v₀/ g
     t₆₀ = 0.86  v₀/g
    Â
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the answer for this part is
      R₄₅ > R₆₀
      t₄₅ < t₆₀
when reviewing the different answers the correct one is C