Question

A man standing on the roof of a house of height h throws one particle vertically downwards and another particle horizontally with the same velocity u. Find the ratio of their velocities when they reach the earth’s surface.
A. 1 : 1
B. 2 : 1
C. 3 : 1
D. 1 : 2

Answer 1

Hint: Use the kinematic equation for displacement of a particle. Find the velocity of the particle thrown downwards when it hits the ground. Then find the vertical velocity of the particle thrown horizontally, when it hits the ground. Its horizontal velocity will remain constant, i.e. u.

Formula used:
$2as={{v}^{2}}-{{u}^{2}}$
W=mgh
$K=\dfrac{1}{2}m{{v}^{2}}$

Complete answer:
It is given that two particles are thrown from the roof of a house. One particle is thrown downwards and the other horizontally with same velocity u. Let the height of the house be h.
We will find the velocities of both particles when they reach the earth’s surface with the kinematic equation $2as={{v}^{2}}-{{u}^{2}}$ ….. (i),
where a, u, v and s are the acceleration, the initial velocity, the final velocity and displacement of the particle.
In this case, both the particles are accelerated due to the gravitational force. Therefore, the acceleration of both particles is g in the downwards direction.
And both the particles have the same displacement, i.e. s = -h.
Let us first analyse the motion of the particle thrown downwards.
The initial velocity is u.
a= -g, s = -h.
Substitute the values in equation (i).
$\Rightarrow 2(-g)(-h)={{v}^{2}}-{{u}^{2}}$
$\Rightarrow {{v}^{2}}={{u}^{2}}+2gh$.
$\Rightarrow v=\sqrt{{{u}^{2}}+2gh}$
This means that the velocity of this particle is $v=\sqrt{{{u}^{2}}+2gh}$.
Let's calculate the final velocity of the particle thrown horizontally.
This particle will have a horizontal as well as vertical motion. We will apply equation (i) to the vertical motion of this particle.
Here, u = 0, s = -h, a = -g and $v={{v}_{y}}$.
 Substitute the values in equation (i).
$\Rightarrow 2(-g)(-h)=v_{y}^{2}+0$
$\Rightarrow {{v}_{y}}=\sqrt{2gh}$.
This means that the vertical velocity of this particle when it hits the ground is ${{v}_{y}}=\sqrt{2gh}$.
The horizontal velocity of this particle will remain the same that is equal to u because there is no force acting on the particle in the horizontal direction.
This means that the horizontal velocity of this particle when it hits the ground is ${{v}_{x}}=u$.
Since ${{v}_{y}}$ and ${{v}_{x}}$ are perpendicular, the net velocity of the particle will be $v=\sqrt{v_{x}^{2}+v_{y}^{2}}$
$\Rightarrow v=\sqrt{{{u}^{2}}+2gh}$.
Therefore, the velocities of both the particles when they hit the ground are the same.
This means that the ratio of their velocities when the particles reach the earth’s surface is 1 : 1.

Hence, the correct option is A.

Note:
We can also find the velocities of the particles with the work energy theorem.
The work energy theorem says that the work done on a particle is equal to the change in kinetic energy of the particle.
Work done by gravitational force is given as W=mgh, where h is the height for which the particle is displaced.
Let us consider the masses of the particles as ${{m}_{1}}$ and ${{m}_{2}}$.
Therefore, the work done on mass ${{m}_{1}}$ is $W={{m}_{1}}gh$.
The change in kinetic energy of this particle will be $\Delta K=\dfrac{1}{2}{{m}_{1}}{{v}^{2}}-\dfrac{1}{2}{{m}_{1}}{{u}^{2}}=\dfrac{1}{2}{{m}_{1}}\left( {{v}^{2}}-{{u}^{2}} \right)$.
And $W=\Delta K$
$\Rightarrow {{m}_{1}}gh=\dfrac{1}{2}{{m}_{1}}\left( {{v}^{2}}-{{u}^{2}} \right)$
$\Rightarrow 2gh={{v}^{2}}-{{u}^{2}}$
$\Rightarrow {{v}^{2}}={{u}^{2}}+2gh$
$\Rightarrow v=\sqrt{{{u}^{2}}+2gh}$.
Similarly, the velocity of the other particle will be $v=\sqrt{{{u}^{2}}+2gh}$.