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:
Answer:
v = 1176.23 m/s
y = 741192.997 m = 741.19 km
Explanation:
Given:
M₀ = 9 Kg (Starting mass)
me = 0.225 Kg/s (Fuel consumption rate)
ve = 1980 m/s (Exhaust speed in relation to rocket, at atmospheric conditions)
v =? for t = 20 s
y =?
We apply the formula:
v = ∫((ve*me)/(M₀ - me*t)) dt - ∫g dt where t ranges from 0 to the specified t
⇒ v = - ve*Ln ((M₀ - me*t)/M₀) - g*t
This leads us to:
v = - 1980 m/s*Ln ((9 Kg - 0.225 Kg/s*20 s)/(9 Kg)) - (9.81 m/s²)(20 s)
Thus, we find:
v = 1176.23 m/s
Next, we utilize the next 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 calculate:
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 presented in the images.
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