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 .
Step 1: Given data:
The stone covers h₁, h₂, and h₃ be height after every 5sec.
As the stone falls freely under gravity;
Initial velocity(u)= 0 m/s
Let v₁ and v₂ be the velocity at the end of height h₁ and h₂ .
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= 10ms−2
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=12×10ms−2×(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=10ms−2×5sec
v1=50ms−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;
h2=50ms−1×5sec+12×10ms−2×(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=50ms−1+10ms−2×5sec
v2=50+50
v2=100ms−1
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=100ms−1×5sec+12×10ms−2×(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 .