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
