Question

A particle is released from rest from a tower of
height 3h. The ratio of the intervals of time to cover
three equal heights h is
​

Answer 1

Answer: Use Gallelio Law of Odd Numbers...

Explanation:

It says in equal time intervals, the distance covered is in ratio of 1:3:5:7:9:11.... If you want a proof let me know, I will send you short tricks and time saving formulae and tips..

Just mark as brainliest

Answer 2

Given:

Initial velocity of particle, u=0

Height of tower, H= 3h

To Find:

Ratio of the intervals of time to cover  three equal heights h

Diagram:

(See attachment)

Solution:

We know that,

• According to second equation of motion for constant acceleration,

$\orange{\boxed{\bf{s=ut+\dfrac{1}{2}at^{2}}}}$

where,

u is initial velocity

a is acceleration

s is displacement

t is time taken

$\rule{190}{1}$

Reference taken here:

All velocities, forces and accelerations acting in upward direction are taken positive.

All velocities, forces and accelerations acting in downward direction are taken negative.

$\rule{190}{1}$

Let the time taken by particle to reach A from O be t₁, time taken by particle to reach B from O be t₂ and time taken by particle to reach C from O be t₃

We have to consider 3 cases:

Case-1 Motion of particle from O to A

On applying second equation of motion on particle, we get

$\longrightarrow\rm{s=ut+\dfrac{1}{2}at^{2}}$

$\longrightarrow\rm{-h=(0)t_{1}+\dfrac{1}{2}(-g)(t_{1})^{2}}$

$\longrightarrow\rm{-h=-\dfrac{g(t_{1})^{2}}{2}}$

$\longrightarrow\rm{h=\dfrac{g(t_{1})^{2}}{2}}$

$\longrightarrow\rm{\dfrac{g(t_{1})^{2}}{2}=h}$

$\longrightarrow\rm{g(t_{1})^{2}=2h}$

$\longrightarrow\rm{(t_{1})^{2}=\dfrac{2h}{g}}$

$\longrightarrow\rm\pink{t_{1}=\sqrt{\dfrac{2h}{g}}}$

Case-2 Motion of particle from O to B

On applying second equation of motion on particle, we get

$\longrightarrow\rm{s=ut+\dfrac{1}{2}at^{2}}$

$\longrightarrow\rm{-2h=(0)t_{2}+\dfrac{1}{2}(-g)(t_{2})^{2}}$

$\longrightarrow\rm{-2h=-\dfrac{g(t_{2})^{2}}{2}}$

$\longrightarrow\rm{2h=\dfrac{g(t_{2})^{2}}{2}}$

$\longrightarrow\rm{\dfrac{g(t_{2})^{2}}{2}=2h}$

$\longrightarrow\rm{g(t_{2})^{2}=4h}$

$\longrightarrow\rm{(t_{2})^{2}=\dfrac{4h}{g}}$

$\longrightarrow\rm{t_{2}=\sqrt{\dfrac{4h}{g}}}$

Case-3 Motion of particle from O to C

On applying second equation of motion on particle, we get

$\longrightarrow\rm{s=ut+\dfrac{1}{2}at^{2}}$

$\longrightarrow\rm{-3h=(0)t_{3}+\dfrac{1}{2}(-g)(t_{3})^{2}}$

$\longrightarrow\rm{-3h=-\dfrac{g(t_{3})^{2}}{2}}$

$\longrightarrow\rm{3h=\dfrac{g(t_{3})^{2}}{2}}$

$\longrightarrow\rm{\dfrac{g(t_{3})^{2}}{2}=3h}$

$\longrightarrow\rm{g(t_{3})^{2}=6h}$

$\longrightarrow\rm{(t_{3})^{2}=\dfrac{6h}{g}}$

$\longrightarrow\rm{t_{3}=\sqrt{\dfrac{6h}{g}}}$

$\rule{190}{1}$

Now, let the time taken by particle to reach B from A be t and the time taken by particle to reach C from B be T

So,

$\longrightarrow\rm{t=t_{2}-t_{1}}$

$\longrightarrow\rm{t=\sqrt{\dfrac{4h}{g}}-\sqrt{\dfrac{2h}{g}}}$

$\longrightarrow\rm\pink{t=\sqrt{\dfrac{2h}{g}}(\sqrt{2}-1)}$

Also,

$\longrightarrow\rm{T=t_{3}-t_{2}}$

$\longrightarrow\rm{T=\sqrt{\dfrac{6h}{g}}-\sqrt{\dfrac{4h}{g}}}$

$\longrightarrow\rm\pink{T=\sqrt{\dfrac{2h}{g}}(\sqrt{3}-\sqrt{2})}$

$\rule{190}{1}$

Now, our required ratio is

$\longrightarrow\rm{t_{1}:t:T}$

$\longrightarrow\rm{\sqrt{\dfrac{2h}{g}}:\sqrt{\dfrac{2h}{g}}(\sqrt{2}-1):\sqrt{\dfrac{2h}{g}}(\sqrt{3}-\sqrt{2})}$

$\longrightarrow\rm\green{1:\sqrt{2}-1:\sqrt{3}-\sqrt{2}}$