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]
We calculate for each angle
θ = 45º
R₄₅ = \frac{ v_{o}^2 \ sin 90}{g}
R₄₅ = v₀² / g
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
the answer for this part is
R₄₅ > R₆₀
t₄₅ < t₆₀
when reviewing the different answers the correct one is C
