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