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Verified answer
V₀ = 20 m/sα = 30°
g = 10 m/s²
V₀ₓ = V₀ cos α
= 20 • ½ √3
= 10√3 m/s
V₀y = V₀ sin α
= 20 • ½
= 10 m/s
a. xₘₐₖₛ = V₀² sin 2α / g
= 20² sin 60° / 10
= 40 • ½√3
= 20√3 m
Maka tₘₐₖₛ = xₘₐₖₛ / Vₓ
= 20√3 / 10√3
= 2 s
b.)
t = 1 s
x = V₀ cos α • t
x = 10√3 • 1
x = 10√3
y = V₀ sin α t - ½ gt²
y = (10 • 1) - (½ • 10 • 1²)
y = 5
Koordinat (10√3 , 5)
c.)
Posisi bola di B , tinggi maksimal
yₘₐₖₛ = V₀² sin² α / 2g
= 20² (sin 30°)² / 2(10)
= 20 • ¼
= 5 m
d.) Jarak maksimum tempuh oleh benda
xₘₐₖₛ = V₀² sin 2α / g
= 20² sin 60° / 10
= 40 • ½√3
= 20√3 m