can some one just check my work plz 1. The problem statement, all variables and given/known data model rocket of mass .250 kg is launched vertically with an engine that is ignited at time= 0, the engine provides 20 N.s impulse by firing for 2 sec . upon reaching its maximum hight the rocket deploys a parachute and then desends vertically to the ground. a) find acceleration during the 2sec firing b) what will be the max hight c) at what time after t=0 will the max hight be reached 2. Relevant equations impulse or change in momentum = F x t ..... sumFy = ma = F*engine* - F *gravity* or mg change in y = v inital x t + 1/2 a t^2 3. The attempt at a solution using first equation---> F t = change p ---> F 2 = 20 --> F= 10 N then second equation F *motor* = ma + mg ---> after you solve for F*motor that is * 10=(.250)a+(.25)(9.81)----> a= 30.19 m/s^2 that was for part A of the question and for part C i found velocity using p=mv final - mv inital ---> 20= (.25)v -0 since it starts from rest ----> velocity = 80 m/s then i used the equation Vf^2 = V init ^2 + 2 a change y ----> 6400= 2(30.19) change y , since v init was 0 it cancels ----> and got change in y = 105.995 m for D i used the formula: delta y =( velocity init ) ( time) + 1/2 (acceleration ) (time^2) ---> 105.995= 1/2 (30.19) (time^2) since velocity init was 0 that part cancels ----> t= 2.65 s, and it would make sense because its after 2 s thanks in advance!