How long does an arrow spend in the air shot at 45 degrees going 220m and hitting at same height it was shot?

Question

An arrow is shot at an angle of 45 degrees above the horizontal. The arrow hits a tree a horizontal distance 220m away, at the same height above the ground as it was shot. Use 9.8m/s for the magnitude of the acceleration due to gravity. someone drops an apple from 6m, how long after the arrow is shot should the apple be dropped in order for the arrow to hit the apple?

Answer 1

range = 220 m
v0 = sqrt(range*g / sin(2*theta)) = 46.433 m/s
Hor. velocity v0x = v0*cos(theta) = 32.833 m/s
t = range/v0x = 6.701 s
See ref. for derivation of 1st equation.

Second question, assuming the arrow hits the apple at ground level:
t(fall) = sqrt(h/2g) = 0.553 s
So drop apple at 6.701-0.553 s after shooting