Question

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 upward 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.

Answer 1

v = 1176.23 m/s

y = 741192.997 m = 741.19 km

Step-by-step 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.

Answer 2

Following are the solution to the given points:

Given:

`\to v_o =1980\ (m)/(s)\\\\`

`\to a=g= -9.8 (m)/(s) \ \text{(Diection downwords )}`

Solution:

Using formula:

`\to V= V_o +at\\\\`

`\to v^2-v^2_0= 2ad \\\\`

For point a:

`\to V= V_o +at\\\\`

`=1980+ (-9.8) (20) \\\\`

`= 1980-196 \\\\ =1784\ (m)/(s)\\\\`

The velocity at the conclusion of a
`20` second is
`\bold{1784\ (m)/(s)}\\\\`.

For point b:

Using formula:

`\to v^2-v^2_0= 2ad \\\\`

`\to d=(v^2_0 -V^2)/(2a)\\\\`

`= ((1980)^2 - (1784)^2)/(2* 9.8) \\\\ =(3920400-3182656)/(19.6)\\\\= (737744)/(17.6)\\\\ =37640\ m \\\\ = 37.64\ km\\\\`

In
`20` seconds, the total distance traveled is
`401737 \ \ m (or \ 401.737\ \ km)`.

For point c:

Please find the attached file.