Answer:
The rocket is motion above the ground for 44.7 s.
Explanation:
The equations for the height of the rocket are as follows:
y = y0 + v0 · t + 1/2 · a · t²
and, after the engine fails:
y = y0 + v0 · t + 1/2 · g · t²
Where:
y = height of the rocket at time t
y0 = initial height
v0 = initial speed
t= time
a = acceleration due to the engines
g = acceleration due to gravity
First, let´s calculate the time the rocket is being accelerated by the engines:
(The center of the framer of reference is located at the ground, y0 = 0).
y = y0 + v0 · t + 1/2 · a · t²
1190 m = 0 m + 79.6 m/s · t + 1/2 · 4.10 m/s² · t²
0 = 2.05 m/s² · t² + 79.6 m/s · t - 1190 m
Solving the quadratic equation:
t = 11.5 s (the other value of t is discarded because it is negative).
At that time, the engines fail and the rocket starts to fall. The equation of the height will be:
y = y0 + v0 · t + 1/2 · g · t²
The initial velocity (v0) will be the velocity acquired during 11.5 s of acceleration plus the initial velocity of launch:
v = v0 + a · t
v = 79.6 m/s + 4.10 m/s² · 11.5 s
v = 126.8 m/s
Now, we can calculate the time it takes the rocket to reach the ground (y = 0) from a height of 1190 m:
y = y0 + v0 · t + 1/2 · g · t²
0 m = 1190 m + 126.8 m/s · t - 1/2 · 9.80 m/s² · t²
Solving the quadratic equation:
t = 33.2 s
Then, the total time the rocket is in motion is (33.2 s + 11.5 s) 44.7 s
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