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**can some one just check my work plz**

## Homework Statement

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

## Homework 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

## 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!

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