Answer:
a = 15.68 [m/s²]; v = 112.92 [m/s]
Explanation:
To solve this problem we must use the combination of two interesting topics of physics, the principle of energy conservation and then the application of kinematics.
The principle of Energy Conservation tells us that kinetic energy is transformed into potential energy or vice versa. For this particular problem, we must imagine the rocket at a height above 650 [m], suddenly this rocket falls to the ground. We will propose the reference point of potential energy at ground level, at this point the potential energy is equal to zero.
[tex]E_{pot}=E_{kin}[/tex]
[tex]E_{pot}=m*g*h\\E_{kin}=\frac{1}{2}*m*v^{2}[/tex]
where:
m = mass of the rocket [kg]
g = gravity acceleration = 9.81[m/s²]
v = velocity of the rocket [m/s]
m*g*h = 0.5*m*v²
9.81*650 = 0.5*v²
v = √((6376.5)/0.5)
v = 112.92 [m/s]
Now using the following kinematic equation we have:
[tex]v_{f} = v_{o} + (a*t)[/tex]
where:
Vf = final velocity = 112.92 [m/s]
Vo = initial velocity = 0 (at the begining the rocket is at rest)
a = rocket acceleration [m/s²]
t = time = 7.2 [s]
Note: in the kinematics equation the positive sign of acceleration means that the rocket accelerates in the direction of motion, i.e. activates its thrusters to descend and descend to 650 [m] in 7.2 [s].
112.92 = 0 + (a*7.2)
a = 15.68 [m/s²]
