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?

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

The value of T must be 6.75 s

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.

because if the tool that you are using is not balanced it there would be false mass.