Answer:
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.
