A couple of physics problems. Anything is accpetable?

Question

1. Jules Verne in 1865 proposed sending men to the Moon by firing a space capsule from a 220-m-long cannon with final speed of 10.97 km/s. What would be the unrealistically large acceleration experience by the space traverlers during the launch.( A human can stand an acceleration of 15g for a short time.)
2. An aircraft has a lift-off speed of 120 km/h. What minimum constant acceleration does this require if the aircraft is to be airborn after a takeoff run of 240m.
3. A train is traveling down a track at 20m/s when the engineer applies the brakes, resulting in an acceleration of -1.0m/s squared as long as the train is in motion. How far does the train move during a 40s interval starting at the instant the brakes were applied?
4. A mailbag is released from a helicopter that is descending steadily at 1.5m/s. After 2.00s, (a) what is the speed of mailbag (b) how fart is it below the helicopter (c) what would be the answers to A and B if the helicopter was going up at speed of 1.5m/s

Answer 1

1. v = at or t = v/a
s = (1/2)at^2 = (1/2)a(v/a)^2
sa = (1/2)v^2

.22(a) = (1/2)(10.97)^2
a = 273.5 km/s^2

2. Using the above: sa = (1/2)v^2
v = 120 km/h = 120(1000/3600) m/s
v = 33.33333 m/s

240(a) = (1/2)*(33.3333)^2
a = 2.3148 m/s^2

3. s= v0t + (1/2)at^2
s = (20)(40) +(1/2)(-1)(40)^2
s = 0 .... the train stops before the 40 seconds are up

v = v0 + at
0 = 20 + (-1) t
t = 20 seconds
So the trains stops after 20 seconds

s = v0t + (1/2)at^2 = 20(20) - (1/2(20)^2
s = 200 m
The train moves 200 meters in the 40 seconds after the brakes have been applied but it stops at 20 seconds.

4.
a. v = v0 + at = 1.5 + 9.8(2)^2
v = 40.7 m/s downward

b. s = v0t + (1/2)at^2
s = 1.5(2) + (1/2)(9.8)(2)^2
s = 22.6 meters
It is 22.6 meters below the helicopter

c. v = v0 + at = -1.5 + 9.8(2)^2 = 37.7 m/s downward

s = v0t + (1/2)at^2
s = -1.5(2) + (1/2)(9.8)(2)^2
s = 16.6 meters beneath the helicopter