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Question

A stone falls freely under gravity. It covers distances h1, h2, h3 in the first 5 seconds , the next 5 seconds and the next 5 seconds respectively. What is the relation between h1, h2 and h3 .

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Solution

Step 1: Given data:

The stone covers h, h, and h be height after every 5sec.

As the stone falls freely under gravity;

Initial velocity(u)= 0 m/s

Let v and v be the velocity at the end of height h and h .

Step 2: Formula used:

v=u+at (First equation of motion)

s=ut+12at2 (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+12gt2

When the body falls freely,

v=gt

and, h=12gt2

g= 10ms2

Step 3: Calculation of the velocity of the stone at h1

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

h1=12×10ms2×(5sec)2

h1=5×25

h1=125m

Let v1 be the velocity of the stone after the first 5sec .

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

v1=10ms2×5sec

v1=50ms1

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;

h2=50ms1×5sec+12×10ms2×(5sec)2

h2=250+5×25

h2=250+125

h2=375m

Let v2 be the final velocity of the stone after the next 5sec, then

Using the first equation of motion under gravity;

v2=50ms1+10ms2×5sec

v2=50+50

v2=100ms1

Step 5: Finding the final height h₃

The final velocity that is v2

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

h3=100ms1×5sec+12×10ms2×(5sec)2

h3=500+5×25

h3=500+125

h3=625m

Step 6: Calculation of the relation between h1,h2 and h3:

We have

h₁=125m

h₂= 375m

and h₃= 625m

h₂= 3 h₁ and h₃= 5 h₁

Thus, h1=h23 and h1=h35

Hence,

The relation between h₁, h₂, and h₃ is h1=h23=h35 .


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