Question

A stone falls freely under gravity. It covers distances h_1, h_2, h₃ in the first 5 seconds , the next 5 seconds and the next 5 seconds respectively. What is the relation between h_1, h₂ and h_{3 .}

Answer 1

Step 1: Given data:

The stone covers $\text{h₁, h₂,}$ and $\text{h₃}$ be height after every $\text{5sec}$.

As the stone falls freely under gravity;

Initial velocity($u$)= $\text{0 m/s}$

Let $\text{v₁}$ and $\text{v₂}$ be the velocity at the end of height $\text{h₁}$ and $\text{h₂}$ .

Step 2: Formula used:

$\text{v=u+at}$ (First equation of motion)

$s=ut+\frac{1}{2}a{t}^2$ (Second equation of motion)

The equation of motion under gravity for a body moving towards the surface of the earth;

$v=u+gt$

and, $h=ut+\frac{1}{2}g{t}^2$

When the body falls freely,

$v=gt$

and, $h=\frac{1}{2}g{t}^2$

g= $10m{s}^{-2}$

Step 3: Calculation of the velocity of the stone at $h_1$

Using the second equation of motion under gravity for a freely falling body;

$h_1=\frac{1}{2}\times 10m{s}^{-2}\times {\left(5sec\right)}^2$

$h_1=5\times 25$

$h_1=125m$

Let $v_1$ be the velocity of the stone after the first $5sec$ .

Then, using the first equation of motion under gravity for a freely falling body;

$v_1=10m{s}^{-2}\times 5sec$

$v_1=50m{s}^{-1}$

Step 4: Finding the final velocity for height h₂

The final velocity after 5sec will be the initial velocity(u₁) of the stone for the next 5 sec,

Now,

Using the second equation of motion under gravity for a freely falling body;

$h_2=50m{s}^{-1}\times 5sec+\frac{1}{2}\times 10m{s}^{-2}\times {\left(5sec\right)}^2$

$h_2=250+5\times 25$

$h_2=250+125$

$h_2=375m$

Let $v_2$ be the final velocity of the stone after the next 5sec, then

Using the first equation of motion under gravity;

$v_2=50m{s}^{-1}+10m{s}^{-2}\times 5sec$

$v_2=50+50$

$v_2=100m{s}^{-1}$

Step 5: Finding the final height h₃

The final velocity that is $v_2$

Then, again Using the second equation of motion under gravity for a freely falling body;

$h_3=100m{s}^{-1}\times 5sec+\frac{1}{2}\times 10m{s}^{-2}\times {\left(5sec\right)}^2$

$h_3=500+5\times 25$

$h_3=500+125$

$h_3=625m$

Step 6: Calculation of the relation between $h_1,h_2$ and $h_3$:

We have

h₁=125m

h₂= 375m

and h₃= 625m

h₂= 3 h₁ and h₃= 5 h₁

Thus, $h_1=\frac{h_2}{3}$ and $h_1=\frac{h_3}{5}$

Hence,

The relation between h₁, h₂, and h₃ is $h_1=\frac{h_2}{3}=\frac{h_3}{5}$ .