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?

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

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

To determine the greatest distance that a man can throw a stone given a height H, we'll need to consider the laws of projectile motion. The basic principles of projectile motion state that the horizontal and vertical components of motion can be considered independently.

The horizontal motion of the stone is not affected by the height from which it is thrown. Therefore, the distance the stone travels horizontally will depend solely on the initial horizontal velocity and the time of flight. We can use the equation:

d = v * t,

where 'd' is the horizontal distance, 'v' is the horizontal velocity, and 't' is the time of flight.

However, the vertical motion will be influenced by the initial vertical velocity, the height from which the stone is thrown, and the acceleration due to gravity. The vertical distance can be calculated using the equation:

H = (1/2) * g * t^2 + v₀ * t,

where 'H' is the height, 'g' is the acceleration due to gravity (approximately 9.8 m/s²), 't' is the time of flight, and 'v₀' is the initial vertical velocity.

Since we are interested in finding the greatest distance, we need to determine the launch angle that maximizes the horizontal distance. It is well-known that the optimal launch angle for maximum horizontal range is 45 degrees (assuming symmetrical projectile motion).

Therefore, to find the greatest distance, we need to calculate the time of flight at a launch angle of 45 degrees and substitute it into the equation for horizontal distance:

t = 2 * (v₀ * sin(θ)) / g,

where 'θ' is the launch angle and 'v₀' is the initial vertical velocity.

From the equation above, we can see that changing the height from which the stone is thrown (H) does not affect the horizontal distance. Therefore, regardless of the height, the greatest distance the man can throw the stone will be the same as long as the initial launch conditions (velocity and angle) remain constant.