Answer:
The time interval during which the rocket engine provides upward acceleration is 2.1 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 burnout:
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 (downward)
The velocity of the rockey is given by this equation:
v = v0 + a · t (v0 = 0 because the rocket is launched from rest)
v = a · t
and after burnout:
v = v0 + g · t
Where v = velocity at time t
We know that when the altitude is 64 m the velocity is 60 m/s. Then let´s use the following equation system:
y = y0 + v0 · t + 1/2 · a · t² (y0 and v0 = 0)
v = a · t
Then:
64 m = 1/2 · a · t²
60 m/s = a · t
a = 60 m/s / t
Replacing "a = 60m/s / t" in the equation of height:
64 m = 1/2 ·( 60m/s / t) · t²
64 m = 30 m/s · t
t = 64 m / 30 m/s
t = 2.1 s
Then, the time interval during which the rocket engine provides upward acceleration is 2.1 s
