The free falling man and his savior suitcase

The man starts falling down from a height H = 100m. For the first 20m, he does not travel horizontally at all. His horizontal jounrney begins at 80m. The crux of the solution is he needs to cover the horizontal distance of 3m in the same time that he has to reach the ground (worst case scenario).

The first step is the calculation of time required to reach the ground

The formula we are concerned with here is: s = ut + 1/2a(t)^2; where u = 0 and g = 10. This gives us t = √(2s/10)

The time he will take to fall from 100m to 0m is t1 = √(2 x 100/10) = 4.47 s

The time taken to fall from 100m to 80m is t2 = √(2 x 20/10) = 2 seconds

Time he has to work with (t) t = t1 - t2 = 4.72 -2 = 2.47 seconds

He needs to travel 3m horizontally in 2.47 seconds.

The second part is solved by using the Law of Conservation of Momentum

(M x V) = (m x v) where the uppercase letters denote the man and lowercase denote the suitcase.

'V' can be calculated using the distance 3m and time 2.47 seconds. V = 3/2.47 = 1.21 m/s ( this the velocity the man needs to have)

Substituting this in the momentum equation, we get 70 x 1.21 = 2 x v

v = 42.35 m/s or 42.4 m/s

This is the minimum velocity with which he will need to throw the suitcase to save himself.

$\textbf{\underline{Assumptions:}}$

Constant acceleration in the vertical direction "g"

No Drag Forces

Bag is thrown at an instant ( ignoring the unrealistic forces involved to achieve this )

$\textbf{\underline{Analysis:}}$

Using a standard coordinate system fixed to the ground we can determine the vertical component of velocity $v_y$ from Conservation of Energy:

$\displaystyle (M+m)gH = \frac{1}{2}(M+m){v_y}^2 + (M+m)gh$

$\displaystyle {v_y}^2 = 2g(H-h)$

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

With the substitution $\displaystyle v_y = \frac{dh}{dt}$ the differential equation may be solved by separation of variables and integration for the remaining duration of the fall given any height $h$ above the ground.

$\displaystyle -\sqrt{\frac{2}{g}} \int \limits_h^0 \frac{dh}{\sqrt{H-h}} = \int \limits_0^t dt$

$\displaystyle t = \sqrt{\frac{2}{g}} \left( \sqrt{H} - \sqrt{H-h} \right)$

Next, apply Conservation of Momentum at the instant the man throws the bag of mass $m$ to determine the man's horizontal component of velocity $v_x$ . The total momentum in the $x$ direction is initially zero and must be conserved since no external forces are acting on the system ( Man & bag) in the $x$ direction.

$\displaystyle p_{before} = p_{after}$

$\displaystyle 0 = M v_x + m v_{bag}$

$\displaystyle v_x = -\frac{m}{M} v_{bag}$

It can be seen that in order for the man to obtain a positive component of velocity in the $x$ direction, the bag must be thrown to the negative $x$ direction with magnitude $v_{bag}$ . Assuming that is the case, the distance traveled by the man in the positive $x$ direction in time $t$ is given by:

$\displaystyle x = v_x t$

$\displaystyle x = \frac{m}{M} v_{bag} t$

$\displaystyle v_{bag} = \sqrt{\frac{g}{2}} \frac{M}{m} \frac{x}{\sqrt{H} - \sqrt{H-h} } = 42.5 \, \frac{\text{m}}{{\text{s}}}$