A rocket moves upward, starting from rest with an acceleration of 34.4 m/s2 for 4.40 s. It runs out of fuel at the end of the 4.40 s but does not stop. How high does it rise above the ground?

Answer 1

after 4.40s, the height is

34.4*4-4.9*4.4^2 = 42.74m

At that point, its velocity is (34.4-9.8)*4.4 = 108.24m/s

So, now we have the ballistic trajectory of the height

42.7 + 108.24t - 4.9t^2

which has its maximum value at t=11.04

Now just plug that in to find the max height.

Answer 2

Steve, think you're a bit off on the height at the end of the burn. SB 331m. And I would subtract out g, the problem gives a flat 34.4 so v at that time is 151m/s.
In the second stage v goes from 151 to zero under influence of gravity so y = v^2/2a = 151^2/9.8*2 = 1169m. All told 1500m.

Answer 3

Sorry would NOT subtract out g in part 1