An artillery shell is fired at an angle of 62.7
above the horizontal ground with an initial
speed of 1520 m/s.
the acceleration of gravity is 9.8 m/s2 .
find the total time of flight of the shell,
neglecting air resistance.
answer in units of min

This is a motion problem. For bodies in a projectile motion, several equations have already been derived and simplified to account for the behavior of these bodies. 
For the time of flight, the equation is

t = (2*v*sin(A))/g

where 
v = initial velocity of the body
A = angle of the initial velocity
g = acceleration due to gravity (9.81 m/s^2)
t = time of flight

this formula can also be used if the air resistance can be neglected.

therefore, substituting the values, we get

t = 275.65 s or 4.59 min

I'm not sure if this is what you're asking for but i'll give it a shot anyway:
1. Your friend jumps, then lands due to gravity.
2. Someone pushes onto an ice rink, and continues to move due to (i think) friction (?).
3. You're in a school bus and you stand in the middle of the aisle, jump, and land where you had previously because you're at the same speed as the bus.