Find the total time of flight of the shell, neglecting air resistance.

Question

An artillery shell is fired at an angle of 45 ? above the horizontal ground with an initial speed of 1710 m / s. The acceleration of gravity is 9 . 8 m / s 2 . Find the total time of flight of the shell, neglecting air resistance. Answer in units of min.  Then find its horizontal range, neglecting air re-sistance.
Answer in units of km.

Answer 1

At an angle of 45°, the x and y components of velocity are both √2 / 2 times the speed (since the sin and cos of 45° are both √2 / 2).

 

So the initial vertical component of the velocity is 1710 * √2 / 2 = 1209.15 m/s.  This is also the initial horizontal component of velocity.

 

The shell will rise until the vertical component of velocity is 0.  The velocity will decrease at a steady rate of 9.8 m/s per second.  So it will take 1209.15 / 9.8 = 123.4 seconds for the shell to stop rising.  Because of symmetry, it will take the same amount of time for it to fall to the ground again.  So the total flight time will be 246.8 seconds.

 

During that time, the horizontal component of velocity will remain unchanged, at 1209.15 m/s.  So the horizontal range will be 1209.15 m/sec * 246.8 sec = 298418.22 m = 298.4 km.