Question

A man of mass $M$ having a bag of mass $m$ slips from the roof of a tall building of height $H$ and starts falling vertically. When at a height $h$ from the ground, he notices that the ground below him is pretty hard, but there is a pond at a horizontal distance ${\text{x}}$ from the line of fall. In order to save himself, he throws the bag horizontally (with respect to himself) in the direction opposite to the pond. Calculate the minimum horizontal velocity imparted to the bag so that the man lands in the water. If the man just succeeds to avoid the hard ground, where will the bag land?

Answer 1

To solve this question, we have to use the theorem of conservation of momentum and the kinematic equations of motion in order to determine the velocity of the bag. Then, on conserving the position of the centre of mass of the system in the horizontal direction will get the final position of the bag.

Formula used: In this solution we will be using the following formula,
$\Rightarrow {v^2} - {u^2} = 2gy$ where $v$ and $u$ are the final and initial velocity, $g$ is the acceleration due to gravity and $y$ is the displacement.

Complete step by step solution:
The displacement covered by the man when he is at a height $h$ from the ground is
$\Rightarrow y = H - h$ (1)
The vertically downward speed of the man at this point is given by
$\Rightarrow {v^2} - {u^2} = 2gy$
As the man slips from the roof, his initial velocity is $u = 0$ . Also substituting (1) in the above equation we get
$\Rightarrow {v^2} = 2g\left( {H - h} \right)$
$\Rightarrow v = \sqrt {2g\left( {H - h} \right)}$ (2)
Now, let the minimum horizontal velocity imparted to the bad by the man be $U$ . Also, let us consider that the man gains a velocity of $V$ in the opposite direction.
As there is no external force on the (man + bag) system in the horizontal direction, so from the conservation of momentum in the horizontal direction, we have
$\Rightarrow mu - MV = 0$
$\Rightarrow MV = mU$
So we get the velocity of the man in the horizontal direction as
$\Rightarrow V = \dfrac{{mU}}{M}$ (3)
As the velocity $U$ of the bag is minimum, so is the velocity $V$ of the man. Hence, the velocity $V$ is just sufficient to make the man reach the pond which is situated at a distance of ${\text{x}}$ . So, the time required for the man to reach the pond is given by
$\Rightarrow t = \dfrac{x}{V}$
From (3) by substituting we get
$\Rightarrow t = \dfrac{{Mx}}{{mU}}$ (4)
As the man falls into the pond in this time, this is also equal to the time required for the man to reach the ground in the vertical direction. As the height of the man is $h$ above the ground, so the displacement of the man to reach the ground is equal to $h$ . Hence, from the second kinematic equation of motion we have
$\Rightarrow h = vt + \dfrac{1}{2}g{t^2}$ (5)
Substituting (2) and (4) in (5) we get
$\Rightarrow h = \sqrt {2g\left( {H - h} \right)} \left( {\dfrac{{Mx}}{{mU}}} \right) + \dfrac{1}{2}g{\left( {\dfrac{{Mx}}{{mU}}} \right)^2}$
Multiplying with ${U^2}$ both the sides
$\Rightarrow h{U^2} = \sqrt {2g\left( {H - h} \right)} \left( {\dfrac{{Mx}}{m}} \right)U + \dfrac{1}{2}g{\left( {\dfrac{{Mx}}{m}} \right)^2}$
On rearranging
$\Rightarrow h{U^2} - \sqrt {2g\left( {H - h} \right)} \left( {\dfrac{{Mx}}{m}} \right)U - \dfrac{1}{2}g{\left( {\dfrac{{Mx}}{m}} \right)^2} = 0$ (6)
So we get a quadratic equation in the terms of $U$ .
The discriminant of this equation is
$\Rightarrow D = {b^2} - 4ac$
$\Rightarrow D = {\left( { - \sqrt {2g\left( {H - h} \right)} \left( {\dfrac{{Mx}}{m}} \right)} \right)^2} - 4h\left( { - \dfrac{1}{2}g{{\left( {\dfrac{{Mx}}{m}} \right)}^2}} \right)$
On solving, we get
$\Rightarrow D = 2g\left( {H - h} \right){\left( {\dfrac{{Mx}}{m}} \right)^2} + 2gh{\left( {\dfrac{{Mx}}{m}} \right)^2}$
On simplifying the above equation, we get
$\Rightarrow D = 2gH{\left( {\dfrac{{Mx}}{m}} \right)^2}$
From the quadratic formula, we have
$\Rightarrow U = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ (7)
Putting the values of the coefficients and the discriminant from (6) and (7) we get
$\Rightarrow U = \dfrac{{ - \left[ { - \sqrt {2g\left( {H - h} \right)} \left( {\dfrac{{Mx}}{m}} \right)} \right] \pm \sqrt {2gH{{\left( {\dfrac{{Mx}}{m}} \right)}^2}} }}{{2h}}$
$\Rightarrow U = \dfrac{{\sqrt {2g\left( {H - h} \right)} \left( {\dfrac{{Mx}}{m}} \right) \pm \sqrt {2gH} \left( {\dfrac{{Mx}}{m}} \right)}}{{2h}}$
Ignoring the negative root we have
$\Rightarrow U = \dfrac{{\sqrt {2g\left( {H - h} \right)} \left( {\dfrac{{Mx}}{m}} \right) + \sqrt {2gH} \left( {\dfrac{{Mx}}{m}} \right)}}{{2h}}$
$\Rightarrow U = \dfrac{{Mx}}{m}\dfrac{{\sqrt {2g} }}{{2h}}\left( {\sqrt {\left( {H - h} \right)} + \sqrt H } \right)$
This is the minimum horizontal velocity of the bag.
Let the bag fall at a horizontal distance of $d$ from the line of fall, opposite to the pond.
Now, as there is no external force on the system in the horizontal direction, the horizontal displacement of the centre of mass of the system is equal to zero. So we have
$\Rightarrow \dfrac{{Mx - md}}{{M + m}} = 0$
$\Rightarrow Mx - md = 0$
So we get
$\Rightarrow md = Mx$
$\Rightarrow d = \dfrac{M}{m}x$
Hence the bag will land at a distance of $\dfrac{{Mx}}{m}$ from the line of fall opposite to the pond.

Note:
The displacement of the bag could also be found by considering the projectile motion of the bag. But that would involve a huge amount of calculation, and at the same time chances of committing mistakes are also more. So we should always prefer the method of the centre of mass.