Respuesta :
Answer:
a) The rocket reaches a maximum height of 737.577 meters.
b) The rocket will come crashing down approximately 17.655 seconds after engine failure.
Explanation:
a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.
1st Stage - Engine
Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ([tex]v[/tex]), measured in meters per second, by using this kinematic equation:
[tex]v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}[/tex] (1)
Where:
[tex]a[/tex] - Acceleration, measured in meters per square second.
[tex]\Delta s[/tex] - Travelled distance, measured in meters.
[tex]v_{o}[/tex] - Initial velocity, measured in meters per second.
If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]a = 2.35\,\frac{m}{s^{2}}[/tex] and [tex]\Delta s = 595\,m[/tex], the final velocity of the rocket is:
[tex]v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}[/tex]
[tex]v\approx 52.882\,\frac{m}{s}[/tex]
The time associated with this launch ([tex]t[/tex]), measured in seconds, is:
[tex]t = \frac{v-v_{o}}{a}[/tex]
[tex]t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }[/tex]
[tex]t = 22.503\,s[/tex]
2nd Stage - Gravity
The rocket reaches its maximum height when final velocity is zero:
[tex]v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})[/tex] (2)
Where:
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]v[/tex] - Final speed, measured in meters per second.
[tex]a[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]s_{o}[/tex] - Initial height, measured in meters.
[tex]s[/tex] - Final height, measured in meters.
If we know that [tex]v_{o} = 52.882\,\frac{m}{s}[/tex], [tex]v = 0\,\frac{m}{s}[/tex], [tex]a = -9.807\,\frac{m}{s^{2}}[/tex] and [tex]s_{o} = 595\,m[/tex], then the maximum height reached by the rocket is:
[tex]v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})[/tex]
[tex]s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}[/tex]
[tex]s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}[/tex]
[tex]s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]s = 737.577\,m[/tex]
The rocket reaches a maximum height of 737.577 meters.
b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:
[tex]s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}[/tex] (2)
Where:
[tex]s_{o}[/tex] - Initial height, measured in meters.
[tex]s[/tex] - Final height, measured in meters.
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]a[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]t[/tex] - Time, measured in seconds.
If we know that [tex]s_{o} = 595\,m[/tex], [tex]v_{o} = 52.882\,\frac{m}{s}[/tex], [tex]s = 0\,m[/tex] and [tex]a = -9.807\,\frac{m}{s^{2}}[/tex], then the time needed by the rocket is:
[tex]0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}[/tex]
[tex]-4.904\cdot t^{2}+52.882\cdot t +595 = 0[/tex]
Then, we solve this polynomial by Quadratic Formula:
[tex]t_{1}\approx 17.655\,s[/tex], [tex]t_{2} \approx -6.872\,s[/tex]
Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.
