Answer:
The speed of the F-4 Phantom upon impact is approximately 214.92 m/s
Explanation:
The given parameter of the motion of the F-4 Phantom jet aircraft are;
The mass of the F-4 phantom rocket, m₁ = 19,100 kg
The mass of the retaining wall, m₂ = 469,000 kg
The velocity of the combined mass of the wall and the F-4 after collision, v₃ = 8.41 m/s
The retainment wall was initially at rest, therefore, v₂ = 0 m/s
Let 'v₁', represent the speed of the F-4 Phantom upon impact
By the principle of conservation of linear momentum, we have;
m₁·v₁ + m₁·v₂ = m₁·v₃ + m₂·v₃ = (m₁ + m₂)·v₃
Plugging in the values, we have;
19,100 × v₁ + 469,000 × 0 = (19,100 + 469,000) × 8.41
∴ v₁ = (19,100 + 469,000) × 8.41/(19,100) = 214.917329843
v₁ ≈ 214.92
The speed of the F-4 Phantom upon impact, v₁ ≈ 214.92 m/s.
