Astomp rocket takes 1.5 seconds to reach its maximum height what was the initial velocity and what was the maximum height ?

Below is the solution:

1) Let the acceleration is a m/s^2 and the velocity after 4 sec is v m/s 
=>By v = u + at 
=>v = 4a (i) 
For 1st phase:- 
=>By s = ut + 1/2at^2 
=>s1 = 0 + 1/2 x a (4)^2 
=>s1 = 8a (ii) 
For 2nd phase:- 
=>By s = vt 
=>s2 = 4a x (9.1-4) 
=>s2 = 20.4a (iii) 
=>By (ii) + (iii):- 
=>s1+s2 = 8a + 20.4a 
=>100 = 28.4a 
=>a = 3.52 m/s^2 
2)(a) By v = u -gt 
=>0 = u - 9.8 x 1.5 
=>u = 14.7 m/s 
(b) By v^2 = u^2 - 2gh 
=>0 = (14.7)^2 - 2 x 9.8 x h 
=>h = 11.03 m

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Iwould choose the same momentum because if they throw a baseball very light and not hard at all then doing it with a bowling ball won’t hurt that worse

Answer: fn = wcos50 = 38.89(9.81)(cos50) = 245.23 n