Answer:
The fireball slows down when moving upward. It speeds up when moving downward. So, its velocity is always changing.
Answer:
Explanation:
Given
acceleration of rocket(a)![=5 m/s^2](https://tex.z-dn.net/?f=%3D5%20m%2Fs%5E2)
At h=1000 m rocket burn out
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
![v^2-0=2\times 5\times 1000](https://tex.z-dn.net/?f=v%5E2-0%3D2%5Ctimes%205%5Ctimes%201000)
![v=\sqrt{10^4}=100 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B10%5E4%7D%3D100%20m%2Fs)
(b) time to reach v=100 m/s
v=u+at
![100=0+5\times t](https://tex.z-dn.net/?f=100%3D0%2B5%5Ctimes%20t)
![t=\frac{100}{5}=20 s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B100%7D%7B5%7D%3D20%20s)
(c)Rocket maximum altitude
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
here u=100 m/s
v=0
![a=9.8 m/s^2](https://tex.z-dn.net/?f=a%3D9.8%20m%2Fs%5E2)
![s=\frac{v^2}{2g}](https://tex.z-dn.net/?f=s%3D%5Cfrac%7Bv%5E2%7D%7B2g%7D)
![s=\frac{100^2}{2\times 9.8}](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B100%5E2%7D%7B2%5Ctimes%209.8%7D)
[tex]s=510.204
Therefore maximum altitude=510.204+1000=1510.204 m
Yes yes ! How right you are ! Truer words were never spoken.
The question is not complete
(c) as the change in the dependent variable is in direct CORRELATION to the change in the independent variable.