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. The ratio of their velocities when they reach the earth’s surface will be

The correct option is 1 : 1.

Let us analyze the motion of both particles using the law of conservation of mechanical energy, which states that in the absence of air resistance, the total mechanical energy of a particle remains constant throughout its flight.

Let the mass of each particle be $m$. Since both particles are thrown from the roof of a house of height $h$ with the same initial speed $u$, their initial mechanical energies are identical.

At the top of the house (initial position):
The potential energy ($U_i$) of each particle is:
$U_i=mgh$
The kinetic energy ($K_i$) of each particle, which depends only on the magnitude of the velocity (speed) $u$ and not its direction, is:
$K_i=\frac{1}{2}m{u}^2$
Thus, the total initial mechanical energy ($E_i$) for both particles is:
$E_i=mgh+\frac{1}{2}m{u}^2$

At the ground level (final position):
Taking the ground as the reference level for potential energy, the potential energy ($U_f$) of each particle becomes:
$U_f=0$
Let $v_1$ be the final speed of the first particle (thrown downwards) and $v_2$ be the final speed of the second particle (thrown horizontally) when they reach the ground.
The final kinetic energy ($K_f$) is:
$K_f=\frac{1}{2}m{v}^2$

By applying the conservation of mechanical energy ($E_i=E_f$):
$mgh+\frac{1}{2}m{u}^2=0+\frac{1}{2}m{v}^2$

Dividing the entire equation by mass $m$:
$gh+\frac{1}{2}$

Multiplying by 2 and solving for $v$:
$v^2=u^2+2gh$
$v=\sqrt{u^2+2gh}$

Since this final speed equation does not depend on the initial angle of projection, both particles will hit the ground with the exact same speed:
$v_1=v_2=\sqrt{u^2+2gh}$

Therefore, the ratio of their velocities (magnitudes) when they reach the earth’s surface is:
$v_1:v_2=1:1$