Answers:
a) [tex]t=0.311 s[/tex]
b) [tex]a=86.847 m/s^{2}[/tex]
c) [tex]y=1.736 m[/tex]
d) [tex]V=17.369 m/s[/tex]
Explanation:
For this situation we will use the following equations:
[tex]y=y_{o}+V_{o}t+\frac{1}{2}at^{2}[/tex] (1)
[tex]V=V_{o} + at[/tex] (2)
Where:
[tex]y[/tex] is the height of the model rocket at a given time
[tex]y_{o}=0[/tex] is the initial height of the model rocket
[tex]V_{o}=0[/tex] is the initial velocity of the model rocket since it started from rest
[tex]V[/tex] is the velocity of the rocket at a given height and time
[tex]t[/tex] is the time it takes to the model rocket to reach a certain height
[tex]a[/tex] is the constant acceleration due gravity and the rocket's thrust
The average velocity of a body moving at a constant acceleration is:
[tex]V=\frac{V_{1}+V_{2}}{2}[/tex] (3)
For this rocket is:
[tex]V=\frac{27 m/s}{2}=13.5 m/s[/tex] (4)
Time is determined by:
[tex]t=\frac{y}{V}[/tex] (5)
[tex]t=\frac{4.2 m}{13.5 m/s}[/tex] (6)
Hence:
[tex]t=0.311 s[/tex] (7)
Using equation (1), with initial height and velocity equal to zero:
[tex]y=\frac{1}{2}at^{2}[/tex] (8)
We will use [tex]y=4.2 m[/tex] :
[tex]4.2 m=\frac{1}{2}a(0.311)^{2}[/tex] (9)
Finding [tex]a[/tex]:
[tex]a=86.847 m/s^{2}[/tex] (10)
Using again [tex]y=\frac{1}{2}at^{2}[/tex] but for [tex]t=0.2 s[/tex]:
[tex]y=\frac{1}{2}(86.847 m/s^{2})(0.2 s)^{2}[/tex] (11)
[tex]y=1.736 m[/tex] (12)
We will use equation (2) remembering the rocket startted from rest:
[tex]V= at[/tex] (13)
[tex]V= (86.847 m/s^{2})(0.2 s)[/tex] (14)
Finally:
[tex]V=17.369 m/s[/tex] (15)
