Explanation:
It is given that,
Height, h = 3.1 m
Initial speed of the rocket, u = 0
Final speed of the rocket, v = 28 m/s
(b) Let a is the acceleration of the rocket. Using the formula as :
[tex]a=\dfrac{v^2-u^2}{2h}[/tex]
[tex]a=\dfrac{(28)^2}{2\times 3.1}[/tex]
[tex]a=126.45\ m/s^2[/tex]
(a) Let t is the time taken to reach by the rocket to reach to a height of h. So,
[tex]t=\dfrac{v-u}{a}[/tex]
[tex]t=\dfrac{28\ m/s}{126.45\ m/s^2}[/tex]
t = 0.22 seconds
(c) At t = 0.1 seconds, height of the rocket is given by :
[tex]h=ut+\dfrac{1}{2}at^2[/tex]
[tex]h=\dfrac{1}{2}\times 126.45\times (0.1)^2[/tex]
h = 0.63 meters
(d) Let v' is the speed of the rocket 0.10 s after launch.
So, [tex]v'=u+at[/tex]
[tex]v'=0+126.45\times 0.1[/tex]
v' = 12.64 m/s
Hence, this is the required solution.
