Let us assume that rocket only runs in initial energy and not using its own to flying.
Also , let upward direction is +ve and downward direction is -ve .
Initial velocity , u = 58.8 m/s .
Acceleration due to gravity , 
.
Final velocity , v - = 0 m/s .
We know , by equation of motion .
Hence, this is the required solution .
Um, this doesn't make any sense. By climbing a hill, you are decreasing your momentum and kinetic energy, so it slows you down. The only positive, is after you have climbed the hill, you have more potential energy, and it will be released once you go down the hill, but you will not be as fast as if you ignored the hill.
Given:
density of air at inlet, 
density of air at inlet, 
Solution:
Now,
(1)
where
A = Area of cross section
= velocity of air at inlet
= velocity of air at outlet
Now, using eqn (1), we get:
= 1.14
% increase in velocity = 
=114%
which is 14% more
Therefore % increase in velocity is 14%
