Respuesta :
Answer:
a) The speed of the rocket after the firing of the engine is 60 meters per second.
b) The rocket will reach a maximum height of 243.543 meters.
c) The rocket will take 8.118 seconds to reach maximum height.
Step-by-step explanation:
a) We assume that model rocket accelerates at constant rate, the equation of motion of the vehicle is:
[tex]v_{1} = v_{o} + a\cdot t[/tex] (Eq. 1)
Where:
[tex]v_{o}[/tex] - Initial speed of the rocket, measured in meters per second.
[tex]a[/tex] - Upward acceleration, measured in meters per square second.
[tex]t[/tex] - Time, measured in seconds.
[tex]v_{1}[/tex] - Maximum speed of the rocket during the ascent, measured in meters.
If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]a = 30\,\frac{m}{s^{2}}[/tex] and [tex]t = 2\,s[/tex] the speed of the rocket is:
[tex]v_{1} = 0\,\frac{m}{s}+\left(30\,\frac{m}{s^{2}} \right) \cdot (2\,s)[/tex]
[tex]v_{1} = 60\,\frac{m}{s}[/tex]
The speed of the rocket after the firing of the engine is 60 meters per second.
b) Since then, the rocket is decelerated due to gravity until maximum height is reached. At first we calculate the height reached by the model rocket just after the firing:
[tex]y_{1} = y_{o}+v_{o}\cdot t + \frac{1}{2}\cdot a\cdot t^{2}[/tex] (Eq. 2)
Where:
[tex]y_{o}[/tex] - Initial height of the rocket, measured in meters.
[tex]y_{2}[/tex] - Final height of the rocket, measured in meters.
If we know that [tex]y_{o} = 0\,m[/tex], [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]a = 30\,\frac{m}{s^{2}}[/tex] and [tex]t = 2\,s[/tex], the final height of the rocket just after the firing is:
[tex]y_{1} = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (2\,s)+\frac{1}{2}\cdot \left(30\,\frac{m}{s^{2}} \right)\cdot (2\,s)^{2}[/tex]
[tex]y_{1} = 60\,m[/tex]
Now we calculate the maximum height reached by the rocket:
[tex]v_{2}^{2}=v_{1}^{2}-2\cdot g\cdot (y_{2}-y_{1})[/tex] (Eq. 3)
Where:
[tex]v_{1}[/tex] - Speed of the rocket right after the firing, measured in meters per second.
[tex]v_{2}[/tex] - Speed of the rocket at maximum height, measured in meters per second.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y_{1}[/tex] - Height of the rocket after the firing, measured in meters.
[tex]y_{2}[/tex] - Maximum height of the rocket, measured in meters.
As next step we clear the maximum height of the rocket:
[tex]v_{1}^{2}-v_{2}^{2} = g\cdot (y_{2}-y_{1})[/tex]
[tex]y_{2}=y_{1}+\frac{v_{1}^{2}-v_{2}^{2}}{2\cdot g}[/tex]
If we get that [tex]y_{1} = 60\,m[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 60\,\frac{m}{s}[/tex], [tex]v_{2} = 0\,\frac{m}{s}[/tex], the maximum height of the rocket is:
[tex]y_{2} = 60\,m+\frac{\left(60\,\frac{m}{s}\right)^{2}-\left(0\,\frac{m}{s} \right)^{2} }{2\cdot \left(9.807\,\frac{m}{s^{2}}\right) }[/tex]
[tex]y_{2} =243.543\,m[/tex]
The rocket will reach a maximum height of 243.543 meters.
c) The total time to reach maximum height is the sum of firing time ([tex]t_{F}[/tex]) and free fall ([tex]t_{FF}[/tex]), all measured in seconds. The free fall time comes from the following formula:
[tex]v_{2} = v_{1} +g\cdot t_{FF}[/tex] (Eq. 4)
[tex]t_{FF} = \frac{v_{2}-v_{1}}{g}[/tex]
If we know that [tex]v_{1} = 60\,\frac{m}{s}[/tex], [tex]v_{2} = 0\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the free fall time is:
[tex]t_{FF} = \frac{0\,\frac{m}{s}-60\,\frac{m}{s} }{-9.807\,\frac{m}{s^{2}} }[/tex]
[tex]t_{FF} = 6.118\,s[/tex]
The time needed for the rocket to reach maximum height is:
[tex]t_{T} = t_{F}+t_{FF}[/tex] (Eq. 5)
[tex]t_{T} = 2\,s+6.118\,s[/tex]
[tex]t_{F} = 8.118\,s[/tex]
The rocket will take 8.118 seconds to reach maximum height.
