free-fall problem. help plz...?

Question

A model rocket is launched straight upward
with an initial speed of 47.8 m/s. It acceler-
ates with a constant upward acceleration of
2.19 m/s2 until its engines stop at an altitude
of 180 m.
a) What is the maximum height reached by
the rocket? Answer in units of m.
b) When does the rocket reach maximum
height? Answer in units of s.
c) How long is the rocket in the air? Answer
in units of s.

Answer 1

♠ at h1=180m altitude the toy’s kinetic energy is
E=E1+E2, where
initial kinetic E1=0.5m*u^2,
initial speed u=47.8m/s, m is toy’s mass,
energy E2=F2*h1, F2=ma being part of hauling force
of toy’s engine to create acceleration a=2.19m/s^2,
(another part is to get better of gravity mg);
♣ after reaching h1 this kin E was spent on further ascent h2 to transform into pot energy mg*h2, that is E=mg*h2,
(a)♥ hence h2=E/(mg) =(0.5m*u^2 + ma*h1)/(mg);
thus max altitude h=h1+h2 = h1 +(0.5u^2 +a*h1)/g =
= 180 +(0.5*47.8^2 +2.19*180)/9.8 =336.80 m
♦ time to reach h1 we get from eqn
h1=u*t1 +0.5a*t1^2 → a*t1^2 +2u*t1 -2*h1=0,
hence t1=(-u +√(u^2 +2a*h1))/a;
♦ time of further ascent h2 is t2=v/g,
where v=u+a*t1 is toy’s speed at h1;
(b)♥ total time to reach vertex is t=t1+t2=
= t1 +(u+a*t1)/g =
= u/g +(1+a/g)*(-u +√(u^2 +2a*h1))/a =
= -u/a +(1/a +1/g)* √(u^2 +2a*h1) =
= -47.8/2.19 +(1/2.19 +1/9.8) *√(47.8^2 +2*2.19*180) =
= 9.144 s;
◙ now its pot energy being mgh will turn into kin energy 0.5m*w^2, where w is its speed just before it hits the ground; thus w=√(2gh);
(c)♥ hence lapse from vertex to ground is
t3=w/g =√(2h/g) =√(2*336.80/g) =8.291 s;
total time in the air =t+t3= 9.144 +8.291 =17.435 s;
click me if confused or w/e;