Respuesta :
Answer:
64.2 m/s
Explanation:
In the x direction:
x = x₀ + v₀ₓ t + ½ at²
400 m = 0 m + v₀ cos (π/5) t + ½ (0 m/s²) t²
t = 400 / (v₀ cos (π/5))
In the y direction:
y = y₀ + v₀ᵧ t + ½ gt²
0 m = 0 m + v₀ sin (π/5) t + ½ (-9.8 m/s²) t²
0 = v₀ sin (π/5) - 4.9 t
t = v₀ sin (π/5) / 4.9
Therefore:
400 / (v₀ cos (π/5)) = v₀ sin (π/5) / 4.9
1960 = v₀² sin (π/5) cos(π/5)
1960 = ½ v₀² sin(2π/5)
3920 / sin(2π/5) = v₀²
v₀ = 64.2 m/s
Answer:
Initial velocity = 423.08m/s
Explanation:
Using formular for Range of projectile
R =V^2 sin2theta/g
Given:
Range=0.4km= 400m
Theta=3.142/5
g= 9.8m/s^2
400= V^2×sin(2×3.142/5)/9.8
400×9.8= V^2Sin 1.26
3920= 0.0219V^2
V^2= 3920/0.0219
V^2= 178995.43
V=sqrt 178995.43
V= 423.08m/s