Maximizing Range 3

After the body falls $H-h$ and immediately before the collision, it has a velocity of $\sqrt{2g(H-h)}$ downward. Immediately after the collision, the body's velocity has the same magnitude, but its direction is entirely horizontal. The range of a body with an initial velocity of $\sqrt{2g(H-h)}$ horizontally which starts $h$ above the ground is $2H\sqrt{\left(1-\frac{h}{H}\right)\frac{h}{H}}$

For a given $H$ , this is maximized when $\left(1-\frac{h}{H}\right)\frac{h}{H}$ is maximized, which happens when $\frac{h}{H} = \boxed{\frac{1}{2}}$

"perfectly elastic collision" !! Ouch!! that must hurt :)

Assume velocity at collision is Vc, the time from collision to hitting ground is t, and the shifted distance is s.

Vc = sqrt( 2g (H-h) )

t = sqrt ( 2h/g )

s = Vc t

s = 2 sqrt( Hh - h^2 )

ds/dh = ( H - 2h ) / sqrt( Hh - h^2 )

at ds/dh = 0

H = 2h

h/H = 1/2

Since the collision is elastic after colliding with the wall all the verticle velocity achieved by the Man will be converted to horizontal velocity

First we will find this velocity

$V^2$ = $0^2$ + 2g(H - h)

Therefore V = $\sqrt {2g (H- h)}$

Now the horizontal distance will be V × time taken by the Man to reach the ground

h = 0t + $\frac {1}{2}$ g $t^2$

So t = $\sqrt {\frac {2h}{g}}$

So distance = $\sqrt {2g (H- h)}$ × $\sqrt {\frac {2h}{g}}$

= 2 $\sqrt {(H - h)h}$

Now it will achieve its maximum value when

h = $\frac {1}{2}$ H

So $\frac {h}{H}$ = $\boxed {0.5}$

Note - to find the Max value of 2 $\sqrt {(H - h)h}$ follow the following method

First assume that h = xH

So 2 $\sqrt {(H - h)h}$ becomes 2H $\sqrt {( x - x^2 )}$

Now the max value of a quadratic is at $\frac {-b}{2a}$ here Max value of x - $x^2$ will be when x = $\frac {-b}{2a}$ = $\boxed {0.5 }$

Instead of undertaking the ordeal of solving for x, an estimated solution can be guessed. If the awning is high there will be little vertical velocity built up to convert into horizontal velocity. If the awning is low, there will be a lot of vertical velocity to convert into horizontal velocity, but, since gravity will make the man resume the fall, the remaining falling distance will be covered very quickly, with no time to take advantage of greater velocity. Therefore one can guess that the best combination of horizontal velocity and remaining falling time is produced by an awning near the midpoint and h/H is close to 0.5.

This problem has a close relative here .

From the equation of motion: v $^{2}$ = u $^{2}$ + 2as, we have, v $^{2}$ = 2g(H-h)

v = $\sqrt{2g(H-h)}$

Since the man undergoes an elastic collision, the momentum will remain constant. Thus, the velocity after collision would be,

v = $\sqrt{2g(H-h)}$

Time taken for the man to land(t) = $\sqrt{\frac{2h}{g}}$

Range(R) = (initial velocity)* (time) = v*t = $\sqrt{2g(H-h)}$ $\sqrt{\frac{2h}{g}}$ = 2 $\sqrt{h(H-h)}$

For range to be maximum,

$\frac{dR}{dh}$ = 0

Thus we obtain,

2( $\frac{1}{2}$ )[h(H-h)] $^{-1/2}$ (H-2h) = 0

Therefore, H - 2h = 0 => H = 2h

So, $\frac{h}{H}$ = 0.5

This is exactly the same as the classic "water squirting from the side of a tank" problem, and as any student of physics knows the range is maximum when the hole is 1/2 way up the tank. I first saw this when I was 12, I think.

Let's call height above the awning $h_1$ , and rename $h=h_2$ , then $H=h_1+h_2$ . The relation between height and time for a free fall is given by $h_1=\frac{gt_1^2}{2}$ . The acquired velocity $gt_1$ is then directed horizontally so that $v_x=gt_1$ . The vertical velocity is reset to 0 so that $h_2=\frac{gt_2^2}{2}$ . During the second part of the fall the horizontal velocity does not change, so that $x=v_x t_2 = gt_1t_2$ .

From these equations it follows that

$\frac{2H}{g}=t_1^2+t_2^2$

$t_2=\sqrt{\frac{2H}{g}-t_1^2}$

$x= gt_1 \sqrt{\frac{2H}{g}-t_1^2}$

At maximum x: $\frac{dx}{dt_1}=0$

$g \sqrt{\frac{2H}{g}-t_1^2} +  gt_1 \frac{-2t_1}{2 \sqrt{\frac{2H}{g}-t_1^2}}=0$

$\frac{2H}{g}-t_1^2 -  t_1 ^2=0$

$\frac{2H}{g}=2t_1^2$

$H=gt_1^2=2h_1$

$h_2=H-h_1=\frac{H}{2}$

$\frac{h_2}{H}=\frac{1}{2}$

From the laws of motion, the horizontal distance "x" travelled after falling to height h and deflection at 45°,

x =2√(H-h)h.

Maximize "x" with respect to "h",

(h/H)=0.5

Answer=0.5

This can be solved without long equations if some reasoning is put to task.

The optimum distance will be achieved from the base of the wall if, after the collision, the horizontal velocity is equal to the average vertical velocity (meaning that the height h and the distance of the crash pad to the wall are equal). Allowing the horizontal and vertical distances travelled after the collision to be equal essentially optimizes the distance gained from the potential energy of falling from H to h. Since there is no initial vertical velocity after the collision, the average velocity is equivalent to the velocity at h/2. The horizontal velocity will be the same as the vertical velocity immediately before the collition, since it is elactic. As I mentioned earlier, this must be equivalent to the average vertical velocity after the collision. Therefore, the vertical velocity immediately before the collision must be equal to the velocity after a fall of height h/2. Thus, H = h/2 + h, and the ratio is 0.5.

I got the answer by chance ¯_(ツ)_/¯