Answer:
a) 51.8 m, b) 27.4 m/s, c) 142 m
Explanation:
Given:
v₀ = 42.0 m/s
θ = 60.0°
t = 5.50 s
Find:
h, v, and H
a) y = y₀ + v₀ᵧ t + ½ gt²
0 = h + (42.0 sin 60.0) (5.50) + ½ (-9.8) (5.50)²
h = 51.8 m
b) vᵧ = gt + v₀ᵧ
vᵧ = (-9.8)(5.5) + (42.0 sin 60.0)
vᵧ = -17.5 m/s
vₓ = 42.0 cos 60.0
vₓ = 21.0 m/s
v² = vₓ² + vᵧ²
v = 27.4 m/s
c) vᵧ² = v₀ᵧ² + 2g(y - y₀)
0² = (42.0)² + 2(-9.8)(H - 51.8)
H = 142 m
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Answer:
4 Km/hr
Explanation:
Data obtained from the question include the following:
Mass of ball (Mb) = 15 Kg
Velocity of ball (Vb) = 20 Km/hr
Mass skater (Ms) = 60 Kg
Velocity of skater (Vs) = 0 Km/hr
Velocity after collision (V) =..?
Thus, we can obtain the velocity after collision as follow:
MbVb + MsVs = V(Mb + Ms)
(15 x 20) + (60 x 0) = V (15 + 60)
300 + 0 = 75V
300 = 75V
Divide both side by 75
V = 300/75
V = 4 Km/hr.
Therefore, the velocity after the collision is 4 Km/hr.
Answer:
the answer to the question is calcium