00:01
What is the given question? two identical objects.
00:03
A and b fall from rest from different heights to the ground and feel no appreciable air resistance.
00:11
If object b takes twice as long as object a to reach the ground, what is the ratio of the heights from which a and b fell? we are asked to find a ratio of heights.
00:24
They have given relation between the time taken for the two objects to reach the ground.
00:29
Now before solving this problem we have to develop a basic formula which connects the time and the height because in this question we have given only two things.
00:41
One is the height and other is the relation between time.
00:45
So let us see how to do this.
00:47
Suppose consider this is the ground.
00:50
Here i am drawing this is the ground consider this has the ground.
00:57
Now suppose consider this is an object with mass here.
01:03
Suppose initially it is at a height of h.
01:06
Consider this is initially at height of h.
01:12
It is free.
01:13
So initial velocity u equal to 0 because it starts from rest.
01:17
Suppose you are there here and you hold the mass m with your hands and you just dropped it.
01:24
So initially it velocity u equal to 0.
01:27
Now the mass m falls down with acceleration due to gravity that is g.
01:37
Therefore, we have equations of motion.
01:42
We have equations of motion from the physics.
01:49
The three equations of what i am writing, that is, v equal to u plus a .t.
01:57
So this is the first equation of motion.
02:02
So, this is the first basic equation.
02:05
And the next basic equation is v square minus u square equal to 2a .s.
02:12
This is the second basic equation and third equation is s equal to ut plus half a t squared.
02:22
This is the third basic equation for the equation of motion.
02:27
Note one thing, this is for uniform velocity constant acceleration.
02:36
This equation is of motion applicable for constant acceleration.
02:46
Since here the acceleration due to gravity, the g acceleration, constant we can use the sequence of motion now out of these equations of motions from equation 3 let us consider only equation 3 say unboxing it if you take this equation the terms in the equations of motion s equal to distance travel and u equal to initial velocity you all know that initial velocity we equal to final velocity and you know that a equal to acceleration and t equal to time you all know so i am not writing so these are the basic now from this equation three uh by with the help of this problem so this is our physical problem now if you apply this physical problem to in this equation you can write uh yes yes means the distance travel see here the object initially at height h it slowly travels down and reaches the ground at this point so the total distance is as to travel is h so here s equal to h so here for this case s equal to h and also the acceleration a is here nothing but g a equal to g and also here u equal to zero in this case so now if you apply these three into this equation the final equation for this problem becomes h equal to 0 times t plus half g t square therefore we have we get let's a minute we get h equal to half gt square therefore uh yes half gt square so this is the b formula that we have got now uh from the question uh we have finally found the relation between height and the time t which we are asked and now we have to do the problem so i'm going with tender color in the question i am writing the given thing here given if object b takes twice as long as object a that means time taken for b object b equal to twice as long as time taken for a t and we're asked good to find the ratio of heights the ratio of heights we are asked to find and it is given the two identical objects...