Question

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

Answer 1

Let the velocity with which the stone is thrown is v.
Initially K.E. = 1/2 m v^2
Finally P.E. =mgH
Initially K.E = Final P.E
1/2 mv^2 = mgH
v = sqrt(2gH)
This is the maximum velocity with which man can throw.

The maximum distance one can throw = v^2/2g
(Read about projectile motion if you want to know about this formula)

Put the value of v in the above equation and

Answer 2

......find the maximum distance.

Answer 3

hmmm

Answer 4

to max out on distance, the guy must throw the ball at the optimal angle, which is 45deg to the ground. i can show where the 45deg comes from if required.

this means that he throws the ball with a horizontal velocity of sqrt(2gH)*cos(45deg) .

the time take is the same as the time for the ball to rise and fall in the vertical plane.
that follows from v = u + at giving t = (v-u)/g.

specifically, t = 2 * sqrt(2gH)sin45deg/g

working through the distance * time figures gives 4H*sin45*cos45 = 2H.

Answer 5

the key to all of this is knowing the basic equations for motion at constant acceleration, and applying them correctly in the horizontal and vertical axes.

v = u + at (1)
x = ut + (1/2)at^2 (2)
V^2 = u^2 + 2ax (3)

these should just trip off the tongue.

the other equations for energy follow these. eg (1/2)mv^2 follows from (3), and mgh from (2).