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In the first stage of a two-stage rocket, the rocket is fired from the launch pad starting from rest but with a constant acceleration of 3.50m/s² upward. At 25.0 s after launch, the second stage fires for 10.0 s, which boosts the rocket’s velocity to 132.5 m/s upward at 35.0 s after launch. This firing uses up all of the fuel, however, so after the second stage has finished firing, the only force acting on the rocket is gravity. Ignore air resistance.
(a) Find the maximum height that the stage two rocket reaches above the launch pad.
(b) How much time after the end of the stage-two firing will it take for the rocket to fall back to the launch pad?
(c) How fast will the stage two rocket be moving just as it reaches the launch pad?

Respuesta :

jolis1796

Answer:

a) ymax = 3088.564 m

b) t = 38.6 s

c) vf = 246.16 m/s  (↓)

Explanation:

Given

a) In the first stage

y₀ = 0 m

v₁y = 0 m/s

a₁ = 3.50 m/s²

t₁ = 25 s

we get y as follows

y₁ = y₀ + v₁y*t₁ + 0.5*a₁*t₁²

⇒ y₁ = 0 + 0*25 + 0.5*3.5*25² = 1093.75 m

v₂y = v₁y + a₁*t₁ = 0 m/s + (3.50m/s²)*(25 s) = 87.5 m/s

In the second stage

v₃y = 132.5 m/s

t₂ = 10 s

y₁ = 0 m

v₃y = v₂y + a₂*t₂    ⇒  a₂ = (v₃y - v₂y) / t₂

⇒  a₂ = (132.5 m/s - 87.5 m/s) / 10 s = 4.5 m/s²

we get y as follows

y₂ = y₁ + v₂y*t₂ + 0.5*a₂*t₂²

⇒ y₂ = 0 + 87.5*10 + 0.5*4.5*10² = 1100 m

when the only force acting on the rocket is gravity

y₃ = (v₃y)²/(2*g)

⇒  y₃ = (132.5 m/s)²/(2*9.81 m/s²) = 894.814 m

finally

ymax = y₁ + y₂ + y₃ = 1093.75 m + 1100 m + 894.814 m = 3088.564 m

t₃ = v₃y / g = (132.5 m/s) / (9.81 m/s²) = 13.5 s

b) when y = ymax

y = g*t²/2  ⇒  t₄ = √(2*ymax / g)

⇒  t₄ = √(2*3088.564 m / 9.81 m/s²) = 25.09 s

t = t₃ + t₄ = 13.5 s + 25.09 s = 38.6 s

c) We can use the formula

vf = √(2*g*ymax)

⇒  vf = √(2*9.81 m/s²*3088.564 m) = 246.16 m/s  (↓)

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