A rocket is fired straight up It accelerates upward at 35 mss for 35 s then runs out of fuel Ignore any air resistance effects How long is the rocket in the air?

Answer 1

There are three phases to this calculation:

# The initial powered acceleration # The deceleration under gravity to the top of the trajectory # The acceleration under gravity back to the ground Using the equations:

a = 2(s - ut)/t2

a=(v-vo)/t

Phase 1

Acceleration (a) = 35m/s2

Time (t1) = 35s

Velocity(v) = 35 x 35 = 1,225m/s

Distance travelled (S1) = 21,437m

Phase 2

Acceleration (a) = -9.8m/s2

Initial Velocity (Vo) = 1,225m/s

Distance travelled (S2) = 76,562m

Time (t2) = 125s

Phase 3

Acceleration (a) = 9.8m/s2

Distance Travelled (St) = S1 + S2 = 98,000m

Time (t3) = 141.4s

Thus total time = t1 + t2 + t3 = 35 + 125 + 141.4 = 301.4s