Answer:
[tex]s=43.6124\ m[/tex] is the horizontal distance between the rocket and the target when the rocket should be launched.
Explanation:
Given:
Now the acceleration of the rocket:
[tex]a=\frac{F}{m}[/tex]
[tex]a=\frac{20}{0.95}[/tex]
[tex]a=21.052\ m.s^{-2}[/tex]
Since this acceleration is acting against the gravity therefore there will be a lesser effective acceleration of the rocket:
[tex]a'=a-g[/tex]
[tex]a'=21.052-9.81[/tex]
[tex]a'=11.242\ m.s^{-2}[/tex]
Now the time taken by the rocket to reach the height of 33 m:
[tex]h=u.t+\frac{1}{2} \times g.t^2[/tex]
where:
[tex]u=[/tex] initial velocity of the rocket [tex]=0\ m.s^{-1}[/tex]
[tex]33=0+0.5\times 11.242\times t^2[/tex]
[tex]t=2.4229\ s[/tex]
Now the distance covered by the target in the above time when the rocket has reached the height of the target:
[tex]s=v_h.t[/tex]
[tex]s=18\times 2.4229[/tex]
[tex]s=43.6124\ m[/tex] is the horizontal distance between the rocket and the target when the rocket should be launched.