Physics Free Fall.?

Question

A water rocket, launched from the ground, rises vertically with acceleration of 30 m/s^2 for 8.7 s when it runs out of "fuel". Disregarding air resistance, how high will the rocket rise?

Steps Please. Thanks =)

Answer 1

For the booster phase
y(t)=15*t^2
and at y(8.7)=1135 m

the speed during this phase is
v(t)=30*t
v(8.7)=261 m/s
That's really fast, did you mean 3.0 m/s^2?

I will continue using 30 m/s^2 as stated

Once the fuel runs out the rocket has equation of motion of
y(t)=1135+261(t-8.7)-.5*9.81*(t-8.7)^2
and
v(t)=261-9.81*(t-8.7)

set a variable T=t-8.7

When v(T)=0, the rocket reaches apogee
T=261/9.81
T=26.6 seconds

y(T)for T=26.6 is the height
4607 m is the apogee

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