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### Course: UP Class 11 Physics > Unit 3

Lesson 3: Kinematic equations for uniformly accelerated motion- Choosing kinematic equations
- Choosing the best kinematic equation
- Using equations of motion (1 step numerical)
- Using equations of motion (2 steps numerical)
- Kinematic equations: calculations
- Free fall - 2 body solved numerical
- Solving freefall problems using kinematic formulas

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# Free fall - 2 body solved numerical

Let's solve a numerical on 2 free falling bodies. One stone dropped and another one, thrown up, when and where do they meet? Created by Mahesh Shenoy.

## Want to join the conversation?

- how do we know a falling object on earth has a constant acceleration as stated by the narrator??(1 vote)
- A falling object would generally mean a freely falling object on which no force apart from gravity acts to cause acceleration. And acceleration due to gravity is taken as a constant, g=9.8 ms^-2 (here it is taken as 10 for the sake of simplicity).

Well, to be accurate - acceleration due to gravity is actually not constant and varies with Height. But the variation is negligible and hence ignored in high school Physics calculations. You will learn more about this in Class 11.

:) HOPE THAT HELPED!(3 votes)

- What if we are asked to calculate the final velocity(1 vote)
- Final velocity of the
*1st stone*is**10√20 m/s or 20√5 m/s**(As u=0 m/s, a=10 m/s^2 and h=100 m, when we use the 2nd equation to find 't', we get it as √20 s or 2√5 s. Now if we use the 1st equation, we get v=10√20 m/s or 20√5 m/s).

Final velocity of*2nd stone*is**0 m/s**(because when an object is thrown against gravity its final velocity is 0 m/s).

Hope it helps!(1 vote)

- I think that the answer in this video is correct, but the first equation is wrong. The acceleration due to gravity acts in the same direction for both the stones. He got it right because the other side of the equation was also wrong.(0 votes)
- Acceleration due to gravity decreases when thrown up and increases when dropped down.(0 votes)

## Video transcript

- [Instructor] Let's solve a problem on two objects in free fall. Here it is. A stone is dropped from a
height of hundred meters and at the same time, another stone is thrown up
with 50 meters per second from the bottom, find when and where do
the two stones meet. So let's quickly look at what's given and try to make a drawing. A stone is dropped from a
height of hundred meters. So there's one stone that's
dropped from hundred meters. And at the same time, another one is thrown up
with 50 meters per second from the bottom. And we need to find when and where do the two stones meet. So if one stone is falling down, and another one is thrown up, then they could meet somewhere, right? So maybe this stone falls down here, and in that same time, maybe this goes all the way up till here, and maybe meats over there, they're gonna hit each
other at some point. We need to figure out when, that means we need to
figure out after how long after we have thrown and drop them. So time is what we need to calculate. And we need to also figure out where, where exactly, what point is this. So, maybe we need to calculate the height. Either we can calculate this
distance or from the top, anything is fine, we'll calculate the height, let's say. These are two things we need to calculate. So how do we do this? Well, we know that whenever
things are falling, they have a constant acceleration. And as a result, they should obey the
three equations of motion. And so maybe we can use these equations for both these stones, and then see if we can
calculate h and t from them. All right, so let's do this. Let's call this a stone number one. This is our stone number two. Let's make a division because we will have two
equations for both the stones. So let's start with stone number one. Here's stone one. Let's look at what data,
what is given to us, what do we know about this stone? Well, we know it's being dropped. And that means whenever
you're dropping something, it means that it has an
initial velocity zero, so we know u is zero. We know its acceleration, its acceleration is plus 10
meters per second square. And it is plus because this stone is, as it falls down its velocity increases, and as a result, acceleration is positive, if velocity would decrease,
acceleration would be negative. Okay, t is something that we
don't know just call it as t. We need to calculate that. What else? What about its displacement? Well, we are taking this point at which the two stones are meeting as h, that height is h. And the right now the stone
is a h of hundred meters. So in time t when they meet each other, how much how much will be the displacement of that stone from here to here? What will this length be? Well, that is going to be
hundred minus h, right? And so, it's displacement we can write is hundred meters minus h. What else? What about final velocity? We have no clue about its final velocity. We don't care about it. So let's forget about the final velocity. So we have these four things. Now the question is, which equation can we use to connect these four things? So why not give it a shot yourself? Can you try and pause the video and see which equation, which of these three
equations would you use to connect these four quantities? Go ahead, give it a shot. All right, let's see, we wouldn't use the first equation because it has a V in
it and we don't have V. We wouldn't use a third equation for the same reason there is a V in it. That means we have on the one equations, let me just cut those equations out. So we have only one equation, S equals ut plus half at squared. So let me go ahead directly
substitute in this equation and see what we get. So if we substitute S is hundred meters minus h, that equals ut, u is zero. So this term goes to zero, plus half a, a is 10
meters per second squared, times t squared. And if we simplify that, we'll get hundred meters minus h, that will equal half times 10 is five, so we get five meters per second square, oops, no square here, times t squared. And, we cannot solve
this equation further. Simply because there are two things that we don't know, we don't know h, we don't know t, so we can't solve it. So this is where we have
to leave this equation. So what else do we do? Well, how do we calculate h and t? Well, we have the second stone, maybe we'll build a second equation, and maybe that equation will help us. So again, a great idea to pause and see if you can write the initial data and build the equation for
the second stone yourself. Go ahead, give it a try. Alright, so for stone two, let's write the initial data. What is its initial velocity? Now, it's not zero, it is thrown up with 50 meters per second. What about its acceleration? Well, since it's thrown up, it slows down as it goes up, right? And because it's slowing down, its velocity is decreasing
that when its acceleration is negative 10 meters per second squared. What else? What about its displacement? Well, since the stone
was from here to here, it's this man is just going to be h. And time is, let's call it as t. And again, we don't care
about its final velocity. We don't want that at all. So which equation would
we use to connect them? Well, the same equation, for the same reason, right? We don't want to find velocities. And so if we substitute, we'll get S, which is which in our example, in our case, h equals, ut, that's going to be
50 meters per second t plus half a t squared. And if we simplify this, again, you're minus 10 divided by
two will get a minus five, so let me just quickly write that down. So we'll end up with h equals 50 t minus five t squared. And again, we get an equation with two unknowns. And so we can't solve it. But now we have two equations with two unknowns. And you may have already learned in maths that when we have two
equations with two unknowns, we can solve it. How do we solve things like this? Well, we can substitute for
one variable into another. For example, maybe from this equation, we can calculate what t is and
substitute in this equation. Or maybe we can calculate
what h is from one equation, and substitute in another. So you know what, since I already know what h is over here, let me substitute this
value in this equation and see what we end up with. All right, so let me write this equation down. So if I write this equation down, I'll get hundred meters minus h which I'll substitute from this equation. So let me just go ahead
and copy that part, and paste it over here. So that's my h, so let's put that in bracket. Okay, that should equal five, five meters per second
square times t squared. And now look, we only have one variable, only t. And so if we solve this, we can now calculate what t is. And once we calculate what t is, well, we can use these equations and calculate h. So again, would be a great idea to pause and see if you can try this part yourself. Now it's purely mathematics, the physics has ended. Okay, anyways, let's do this. So let me scroll along, because I don't need any of those things. Okay, so if we further simplify, let's open up this bracket, we'll get a negative here. And this will be positive because negative negative
will be positive. So again, if I quickly write this, we get the same thing,
we get a positive here, and everything else remains the same. And now look, we have a five t square here, and we have a five t squared over here. So that just cancels out, you can subtract five
t square on both sides. And we are lucky because the t squares cancel out. Otherwise, it would have
been a quadratic equation, there'll be more steps. But now because the t squares are gone, that means we are left out with hundred meters
minus 50 t equals zero, equals zero. And we can now solve it
in a couple of steps. So if I add 50 t on both sides, then I'll end up with hundred meters equals 50 meters per second times t. And, if we divide 50 meters
per second on both sides, then we get what t is, so the zero cancels, five goes two times, the meter cancels, and we end up with a
two second on this side. So, we found what t is, t equals two seconds. That means, after two seconds of dropping and throwing up that stone, they will meet each other. Now, before we go back up there, let's calculate what h is, how do we do this? Well, now that we know what t is, we can substitute in any
of these two equation, and I'm pretty sure you can do that you just substitute t s two
in any of this equation, and you can calculate h. And if you do that, which I leave it to you to save time, if you do that, you will end up with h equal to 80 meters. And so now let's zoom out and put everything in one frame. Yeah, there it is. And so whenever we have
two objects in free fall, and we're asked to calculate
when or where they meet, what do we do? Well, we build equation for both of them by using the equations of the motion, and then we saw the two equations to calculate the height and the time.