A200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. it accelerates upward at 30.0 m/s2 for 30.0s, then runs out of fuel. ignore any air resistance effects. part a what is the rocket's maximum altitude?

There are two stages to the flight: acceleration stage and deceleration stage.

m₁ = 200 kg, mass of the rocket
m₂ = 100 kg, mass of fuel
a₁ = 30.0 m/s², upward acceleration when burning fuel
Ignore air resistance and assume g = 9.8 m/s².

Acceleration stage:
The rocket starts from rest, therefore the initial vertical velocity is zero.
The distance traveled is given by
s₁ = (1/2)*(30.0 m/s²)*(30.0 s)² = 13500 m

Deceleration stage (due to gravity):
The initial velocity is u = (30.0 m/s²)*(30 s) = 900 m/s
The initial height is 13500 m
At maximum height, the vertical velocity is zero.
Let s₂ =  the extra height traveled. Then
(900 m/s)² - 2*(9.8 m/s²)*(s₂ m) = 0
s₂ = 900²/19.6 = 41326.5 m

The maximum altitude is 
s₁+s₂ = 54826.5 m

54,826.5 m

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward