Two bodies of masses $m_{1}$ and $m_{2}$ fall from heights $h_{1}$ and $h_{2}$ respectively. The ratio of their velocities, when they hit the ground is

h_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{\frac{h_{1}}{h_{2}}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{m_{1} h_{1}}{m_{1} h_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{h_{1}^{2}}{h_{2}^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\sqrt{\frac{h_{1}}{h_{2}}}$
When a body is at rest at height $h,$ it has only potential energy $= m g h$

when the body hits the ground all potential energy is converted to kinetic energy $= \frac{1}{2} m v^{2}$

From conservation of energy, $= \frac{1}{2} m v^{2} = m g h$

so, $v = \sqrt{(2 g h)}$

For body 1, $v_{1} = \sqrt{(2 g h_{1})} . . . . . . . (1)$

For body 2, $v_{2} = \sqrt{(2 g h_{2})} . . . . . . . (2)$

dividing $(1)$ by $(2)$ we get , $\frac{v_{1}}{v_{2}} = \frac{\sqrt{h_{1}}}{\sqrt{h_{2}}}$

Suggest Corrections

Similar questions

Q. Two objects of masses

m_{1}

and

m_{2}

having the same size are dropped simultaneously from heights

h_{1}

and

h_{2}

respectively. Find out the ratio of time they would take in reaching the ground.

Q. Two bodies of different masses

m_{1}

and

m_{2}

are dropped from two different heights

h_{1}

and

h_{2}

. The ratio of the times taken by the two to reach the ground is

Two block of wood of masses $m_{1}$ and $m_{2}$ and each of specific gravity 0.5 are submerged at a depth of $h_{1}$ and $h_{2}$ in
a vessel $(h_{2} > h_{1})$ filled with water, which is accelerated upwards with an acceleration $\frac{g}{2}$ . the difference in
time taken by the blocks to reach the surface, when released with zero velocity is :