The first ball is thrown upwards at a speed of 40 m / s, 1 second later, a second ball thrown upward from the same position with a speed of 47.5 m / s. The high that achieved a second ball when the second ball met with the first ball is ....
A. 55 m
B. 75 m
C. 95 m
D. 125 m
E. 140 m

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Answers

2016-04-11T03:27:39+05:30
V₁ = 40 m/s
v₂ = 47.5 m/s
Δt = 1 s
g ≈ 10 m/s²  (acceleration of gravity)

The equation of motion for first ball:
h(t) = v₁t - gt²/2
the equation of motion for second ball:
h(t) = v₂(t-Δt) - g(t-Δt)²/2
v₁t - gt²/2 = v₂(t-Δt) - g(t-Δt)²/2
v₁t - gt²/2 - v₂t + v₂Δt + gt²/2 + gΔt²/2 - 2gtΔt/2 = 0
t(v
₁ - v₂ - gΔt) = -(v₂Δt + gΔt²/2)
t = (v₂Δt + gΔt²/2) / (v₂ - v₁ + gΔt)
t = (47.5 * 1 + 10 * 1² / 2) / (47.5 - 40 + 10 * 1) = 52.5 / 17.5 = 3 [s]
h₁(t=3s) = v₁t - gt²/2 = 40 * 3 - 10 * 3² / 2 = 120 - 45 = 75 [m]
h₂(t=3s) = v₂(t - Δt) - g(t-Δt)²/2 = 47.5 * (3 - 1) - 10 * (3 - 1)² / 2 = 95 - 20 = 75 [m]
OK, because h₁ = h₂ after  t = 3 s from thrown of first ball. 
Answer: B (75 m)
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