Question

An artillery shell is fired at an angle of 32.2◦ above the horizontal ground with an initial speed of 1610 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

Find its horizontal range, neglecting air resistance.
Answer in units of km

Answer 1

You can do this yourself,

The time of flight is twice the time it takes for the vertical velocity component to be come zero.

The horizontal range is
2 sin32.2 cos 32.2*(Vo^2/g)
= (Vo^2/g) sin 64.4 = ____

Answer 2

then how would you find vertical velocity component?