The greatest height to which a man can throw a stone is H.what will be the greatest distance upto which he can throw the stone?

Answer 1

find v in terms of H for vertical toss
m g H = (1/2) m v^2
v^2 = 2 gH
v = sqrt(2 g H)

max distance at 45 degrees
cos theta = sin theta = .707
speed = sqrt(2 gH)
horizontal speed = .707 sqrt(2 g H)
Vi up = .707 sqrt(2 gH)
speed at top = 0
so
at top
g t = sqrt(2 gH)
t = sqrt(2 H/g)
time in air = 2 t = 2 sqrt (2 H/g)

horizontal speed times time in air =
.707 sqrt(2 gH) * 2 sqrt(2 H/g)
=(1/sqrt2)(2)(2H)
=( 4/sqrt 2) H
= (2 sqrt 2) H or about 2.8 H

Answer 2

Now about that max range at 45 degrees elevation
speed = s
horizontal speed = u = s cos theta
initial vertical speed = Vi = s sin theta

max height
Vi = g t
t = s/g sin theta
time in air = 2 t = (2 s/g)sin theta

range = r = u * 2 t
= s cos theta * 2 (s/g) sin theta
= (2 s^2/g) cos theta sin theta
dr/dtheta = 0 for max
0 = (2 s^2/g)[ -cos^2 theta + sin^2 theta]
or where cos theta = sin theta which is 45 degrees or pi/4 radians where cos theta = sin theta = 1/sqrt 2 = .707