Answer: 90 m/s
Explanation:
For this situation we will use the following equations:
[tex]y=V_{o}t+\frac{1}{2}at^{2}[/tex] (1)
[tex]V=V_{o} + at[/tex] (2)
Where:
[tex]y=200 m[/tex] is the height of the model rocket at 4 s
[tex]V_{o}=10 m/s[/tex] is the initial velocity of the model rocket
[tex]V[/tex] is the velocity of the rocket at 4 s
[tex]t=4 s[/tex] is the time it takes to the model rocket to reach 200 m
[tex]a[/tex] is the constant acceleration due gravity and the rocket's thrust
Firstly, from equation (1) we have to find [tex]a[/tex]:
[tex]200 m=(10 m/s)(4 s)+\frac{1}{2}a(4 s)^{2}[/tex] (3)
[tex]a=20 m/s^{2}[/tex] (4)
Now we have to substitute this value of [tex]a[/tex] in (2):
[tex]V=10 m/s + (20 m/s^{2})(4 s)[/tex] (5)
Finally:
[tex]V=90 m/s[/tex] This is the rocket's final velocity
