The general equation of a parabola is as follows.

#y = (-gx^2)/(2V^2Cos^2 A) + xtan A#

#A# is the angle at which the projectile is fired.

#g# is #9.81 m/s^2#, acceleration due to gravity.

#V# the initial velocity

when #y= 0# we have #x# as the maximum range

To get the answer for this question, we set #x=y# Therefore,

# x = (-gx^2)/(2V^2cos^2 A) + xtan A#,

then we divide both sides by #x# then we have,

#1 = (-gx)/(2V^2cos^2A) + tan A#

#gx/(2V^2cos^2A ) = tan A - 1#

# x = [((sin A)/(cos A) - 1)(2V^2 cos^2 A)]/g#

# x = [(2V^2sin A cos A - 2V^2 cos^2 A)]/g]#

#2sin A cos A = sin 2A#

# x = ( V^2 sin 2A - 2V^2 cos^2 A)/g#

# x =[ V^2( sin 2A - 2cos^2 A)]/g#

then d = rt

to get the time, **t**, we divide x with V

so

# t = [V( sin 2A - 2 cos^2A)]/g#

**I will still update this answer. If you have something to add pls do so. Thanks**