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During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be launched from rest from the earth's surface and is to reach a maximum height of 960 m above the earth's surface. The rocket's engines give the rocket an upward acceleration so it moves with acceleration of 16.0 m/s2 during the time T that they fire. After the engines shut off, the rocket is in free fall. Ignore air resistance. What must be the value of T in order for the rocket to reach the required altitude?

Please use calculus based equations and write out each step and be as clear as possible with what equations you used.

1 Answer

7 votes

Answer:

The value of T must be 6.75 s

Step-by-step explanation:

The equations for the height of the rocket are as follows:

y = y0 + v0 · t + 1/2 · a · t²

and, after the engines run out of fuel:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the rocket at time t

y0 = initial height

v0 = initial velocity

t = time

a = upward acceleration

g = acceleration due to gravity

The velocity (v) of the rocket will be given by the following equations:

v = v0 + a · t (while the engines are firing)

v = v0 + g · t (when the rocket is in free fall)

The height reached after the upward acceleration phase will be:

y = y0 + v0 · t + 1/2 · a · t² (y0 = 0, v0 = 0, t = T, a = 16.0 m/s²)

y = 1/2 · 16.0 m/s² · T²

y = 8.00 m/s² · T²

The velocity reached after the upward acceleration will be:

v = v0 + a · t (v0 = 0, a = 16.0 m/s², t = T)

v = 16.0 m/s² · T

The velocity of the rocket after the engines run out of fuel will be:

v = v0 + g · t

In this case, v0 will be the velocity reached after the upward acceleration

(v = 16.0 m/s² · T). Then:

v = 16.0 m/s² · T + g · t

When the height is maximum (960 m), the velocity of the rocket will be 0. Then:

0 = 16.0 m/s² · T + g · t

Solving for t

- 16.0 m/s² · T / g = t

-16.0 m/s² · T / -9.8 m/s² = t

t = 1.63 · T

Now, replacing in the equation of height after the engines shut off:

y = y0 + v0 · t + 1/2 · g · t²

being:

t = 1.63 · T (time at which the velocity is 0, i.e, the rocket is at max-height)

y0 = 8.00 m/s² · T² (height reached after the upward acceleration phase)

v0 = 16.0 m/s² · T (velocity reached after the upward acceleration phase)

y = 960 m (maximum height)

g = -9.8 m/s²

Then:

960 m = 8.00 m/s² · T² + 16.0 m/s² · T · 1.63 · T - 1/2 · 9.8 m/s² · (1.63 · T)²

960 m = 34.1 m/s² · T² -13.0 m/s² · T²

960 m = 21.1 m/s² · T²

960 m/ 21.1 m/s² = T²

T = 6.75 s

The value of T must be 6.75 s.

User Mark Bonano
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