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

Ithink your answer is d

The answer is d) by requesting advice on science internet sites