A rocket fires from rest with an upward acceleration of 35 m/s2 for 2 seconds. After this time, the engine shuts off and the rocket freely falls back to the earth. The maximum height is 314 and the height the engine shuts off is 70 m. What is the amount of time the rocket is in the air?

Answer 1

the g in this instance is 10 m/s ^2

Answer 2

first step: h = 1/2 at^2 = 1/2 * 35 * 2^2 = 70 m
at this point, the velocity v = 35*2 = 70 m/s
so now, after t more seconds,
v = 70 - 10t
since v=0 at maximum height, we have t=7 when h = 314
so now we are at t=9
from this point after t more seconds, h=314 -5t^2 and h=0 when t=7.92
so total flight time is about 2+7+8 = 17 s

Answer 3

If it accelerates from rest at 35 m/s^2 for 2 seconds it reaches
(1/2) a t^2 = (35/2)(4) = 70 meters
so agree but wonder why they said it.

the engine shuts off at 70 meters
The rocket then coasts up to 314meters
What is initial speed at 70 m?
Vi = a t = 35 * 2 = 70 m/s at 70 meters and 2 seconds
now it coasts up an additional 314-70 = 244meters
v = Vi - 10 t = 70 - 10 t
v = 0 when t = 10 so 12 seconds in the air so far
now it falls 314 meters
314 = (10/2) t^2
t^2 = 62.8
t = 7.92 seconds
so total time = 12 + 7.92 = about 20 seconds

Answer 4

whoops arithmetic !
v = Vi - 10 t = 70 - 10 t
v = 0 when t = 7 so 9 seconds in the air so far
now it falls 314 meters
314 = (10/2) t^2
t^2 = 62.8
t = 7.92 seconds
so total time = 9 + 7.92 = about 17 seconds