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3 years ago

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A "home-made" solid propellant rocket has an initial mass of 9 kg; 6.8 kg of this is fuel. The rocket is directed vertically upw

ard from rest, burns fuel at a constant rate of 0:225 kg=s, and ejects exhaust gas at a speed of 1980 m=s relative to the rocket. Assume that the pressure at the exit is atmospheric and that air resistance may be neglected. Calculate the rocket speed after 20 s and the distance traveled by the rocket in 20 s. Plot the rocket speed and the distance traveled as functions of time.

1 answer:

nlexa [21]3 years ago

6 0

Answer:

v = 1176.23 m/s

y = 741192.997 m = 741.19 km

Explanation:

Given

M₀ = 9 Kg (Initial mass)

me = 0.225 Kg/s (Rate of fuel consumption)

ve = 1980 m/s (Exhaust velocity relative to rocket, leaving at atmospheric pressure)

v = ? if t = 20 s

y = ?

We use the equation

v = ∫((ve*me)/(M₀ - me*t)) dt - ∫g dt where t ∈ (0, t)

⇒ v = - ve*Ln ((M₀ - me*t)/M₀) - g*t

then we have

v = - 1980 m/s*Ln ((9 Kg - 0.225 Kg/s*20 s)/(9 Kg)) - (9.81 m/s²)(20 s)

v = 1176.23 m/s

then we apply the formula

y = ∫v dt = ∫(- ve*Ln ((M₀ - me*t)/M₀) - g*t) dt

⇒ y = - ve* ∫ Ln ((M₀ - me*t)/M₀) dt - g*∫t dt

⇒ y = - ve*(Ln((M₀ - me*t)/M₀)*t + (M₀/me)*(M₀ - me*t - M₀*Ln(M₀ - me*t))) - (g*t²/2)

For t = 20 s we have

y = Ln((9 Kg - 0.225 Kg/s*20 s)/9 Kg)*(20 s) + (9 Kg/0.225 Kg/s)*(9 Kg - 0.225 Kg/s*20 s - 9 Kg*Ln(9 Kg - 0.225 Kg/s*20 s)) - (9.81 m/s²*(20 s)²/2)

⇒ y = 741192.997 m = 741.19 km

The graphs are shown in the pics.

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