B Independence of vertical and horizontal motion

[rudransh verma]

It says when the ball is shot the can is released and they both hit each other at the same height ie they travel same distance down. But that is only possible when the ball starts it’s downward journey the same time as the can starts it’s own. Shooting a ball upward direction will give it some upward velocity.

 

[berkeman]

Sure, but the point of the example is that if you aim at the starting point of the falling mass and shoot at the same instant that the mass is dropped, the projectile hits the falling mass. Can you write the two equations to show that this is true?

 

[rudransh verma]

Sure, but the point of the example is that if you aim at the starting point of the falling mass and shoot at the same instant that the mass is dropped, the projectile hits the falling mass. Can you write the two equations to show that this is true?

But if we aim upward there will always be upward velocity component. If this velocity component is not zero how could they hit each other.
They will hit if they are dropped from the same height one with some horizontal push and one vertical.

 

[berkeman]

Please just write the equations to convince yourself. This is a classic problem, and a good one to work through.

 

[rudransh verma]

Please just write the equations to convince yourself. This is a classic problem, and a good one to work through.

I don’t know what the eqns will be. I guess it has something to do with 2nd eqn of motion. But I can’t reach to any result.

 

[berkeman]

Well, the equation of motion for the falling mass is pretty easy, right?

And the equation of motion for the projectile will have a horizontal and a vertical component. The horizontal velocity is constant (ignoring air resistance), and the vertical velocity component starts out as $V_{y0}$ and is affected by the acceleration of gravity downward.

Use a coordinate system (x,y) with an origin conveniently placed, and write the equation that shows the time and place where the positions of the two objects are the same...

 

[berkeman]

BTW @rudransh verma -- I just wrote out the equations and indeed the projectile does hit the falling mass. I used an x-y coordinate system with the origin at the point where the mass is released, with y increasing downward and x increasing in the direction of the projectile launch point. I then solved for the time for the projectile to pass x=0, and plugged that back into the equations for the vertical motion of the falling mass and the 2-D motion of the projectile.

 

[Lnewqban]

But if we aim upward there will always be upward velocity component. If this velocity component is not zero how could they hit each other.
They will hit if they are dropped from the same height one with some horizontal push and one vertical.

Are you reading the description of Figure 4-10 or 4-11?

 

[rudransh verma]

Are you reading the description of Figure 4-10 or 4-11?

i have problem with 4 11

 

[Lnewqban]

i have problem with 4 11

Thank you.
The use of the phrase “both fall” seems confusing to me.
For values smaller than g (on the surface of the moon for example), the value of h would be smaller.
That happens because both, the ball moving upwards and the can moving downwards are affected by the same acceleration.
The vertical velocity of the ball decreases at the same rate the vertical velocity of the can increases downwards.

 

[rudransh verma]

BTW @rudransh verma -- I just wrote out the equations and indeed the projectile does hit the falling mass. I used an x-y coordinate system with the origin at the point where the mass is released, with y increasing downward and x increasing in the direction of the projectile launch point. I then solved for the time for the projectile to pass x=0, and plugged that back into the equations for the vertical motion of the falling mass and the 2-D motion of the projectile.

Can you show it

 

[berkeman]

Can you show it

Nope, that is for you to do. I gave pretty good instructions to you for how to conveniently set it up. It also helps if you can post using LaTeX for the math equations. There is a good LaTeX Guide linked below the edit window. My last result was the following, which is true because of how I set up the coordinate system:

$$\frac{y_1}{x_1} = tan(\Theta)$$

 

[rudransh verma]

and x increasing in the direction of the projectile launch point.

How? X of projectile will increase in positive x direction.

 

[berkeman]

How? X of projectile will increase in positive x direction.

You can define the coordinate system, origin and the "positive" directions however you want. As long as you write and solve the equations consistent with the coordinate system definitions, it all works out. I chose positive y down and positive x to the left for my convenience in writing my equations. Go ahead and define the coordinate system however you want if it is more comfortable for you.

 

[rudransh verma]

@berkeman please check

 

[berkeman]

Hmm, I'm not able to follow your work. Can you post a sketch of your coordinate system with the following labeled: $x_2, y_1, y_2$ and say in words what your equations represent?

 

[rudransh verma]

Hmm, I'm not able to follow your work. Can you post a sketch of your coordinate system with the following labeled: $x_2, y_1, y_2$ and say in words what your equations represent?

 

[berkeman]

Well, it looks like you are close to doing it correctly, but IMO you are overloading the subscripted variables in a way that is confusing (at least for me). I use the subscript "p" for the projectile and "m" for the mass, so have the variables $x_p(t)$ and $y_p(t)$ to describe the motion of the projectile, and $y_m(t)$ to describe the vertical motion of the mass.

Then the numeric subscripted values can be the starting points of the projectile and mass, and the end point where they meet. I had $x_1$ and $y_1$ as the starting point of the projectile, and $x=0$ and $y_2$ as the point where they collide.

If you are writing up the solution for somebody else to read and grade, it's best to be clear about what each variable is and use a bit of text as you work along to explain why you are writing a particular equation.

 

[Nugatory]

@rudransh verma , you will find it much easier to write your equations in a form that everyone else can follow, and you will get better and more helpful answers, if you use Latex as @berkeman suggested in post #12 above.

 

[rudransh verma]

Well, it looks like you are close to doing it correctly, but IMO you are overloading the subscripted variables in a way that is confusing (at least for me).

I wrote it clearly as much as possible . I did it as you described. But I don’t think this is the answer showing that the two bodies will collide. What does tan have to do with our goal?

 

[berkeman]

What does tan have to do with our goal?

I described how I approached the problem here:

BTW @rudransh verma -- I just wrote out the equations and indeed the projectile does hit the falling mass. I used an x-y coordinate system with the origin at the point where the mass is released, with y increasing downward and x increasing in the direction of the projectile launch point. I then solved for the time for the projectile to pass x=0, and plugged that back into the equations for the vertical motion of the falling mass and the 2-D motion of the projectile.

Since I was able to solve symbolically for the time for the projectile to reach x=0, I plugged that time back into the vertical motion equations and was able to simplify them down to that final tan() equation which was true by the definition of how I set up my coordinates and the starting point of the projectile. Basically simplifying the equations got me to 1=1, so that showed that the two objects did indeed come together at that time and place.

 

[Doc Al]

Just for fun, here's a demo of this exercise (which was called the "monkey and hunter", back in the day):

 

[berkeman]

LOL the MonkeyCam view!

 

[rudransh verma]

@berkeman i am not getting to the answer
I basically have to make x1=x2=0 and y1=y2 after time t.

 

[Doc Al]

It's a bit difficult to follow some of your writing. Looks like you found the time it takes for the projectile to cover the horizontal distance. Good!

And you found the vertical position of the projectile at that time. Good.

So, what's the vertical position of the can at that time? (Express its initial position in terms of the variables that you've been using.)

 

[sysprog]

But if we aim upward there will always be upward velocity component. If this velocity component is not zero how could they hit each other.
They will hit if they are dropped from the same height one with some horizontal push and one vertical.

If the line-of-sight point of aim is the starting position of the can, but instead of the can being released/dropped when the projectile leaves the barrel, the can is not dropped/released, then the projectile will still arrive at some point directly below (lower than) the aim point, but the aim point will not travel downward to coincide with the projectile impact point.

If you [drop a bullet] and [fire one horizontally] at the same time, both will hit the ground at the same time.

Please consider this:

The inside scale is the cosine of the outside scale. The 4-digit cosine of 45 degrees is .7071. That means that gravity is operating on the projectile for only about 71% of its 'ground distance'.

 

[A.T.]

Just for fun, here's a demo of this exercise (which was called the "monkey and hunter", back in the day):

Also relevant to this trick:

 

[A.T.]

But if we aim upward there will always be upward velocity component. If this velocity component is not zero how could they hit each other.

A simple way to see this, is to use a free falling reference frame, where you effectively have zero g. So the target remains static and the projectile moves along a straight line. Therefore it always hits the target it is aimed at, regardless of the inclination.

 

[rudransh verma]

So, what's the vertical position of the can at that time?

y1=-1/2gt^2

 

[Doc Al]

y1=-1/2gt^2

That's how far it falls. But where did it start from? What is the can's initial position?

 

[rudransh verma]

That's how far it falls. But where did it start from? What is the can's initial position?

Origin

 

[Doc Al]

Origin

Nope. The ball started at the bottom (ground) level, but the can did not. What's the initial height of the can above the ground level?

 

[rudransh verma]

Nope. The ball started at the bottom (ground) level, but the can did not. What's the initial height of the can above the ground level?

I took the starting point of can as origin

 

[Doc Al]

I took the starting point of can as origin

You are finding the position above ground level for both, so measure everything from ground level.

 

[rudransh verma]

@Doc Al Now what?

 

[Doc Al]

@Doc Al Now what?

Again, it's a bit difficult to understand what you've written. Please type out your equations here, not handwritten on paper.

(A) Start here: In terms of the horizontal distance traveled (I'd call it "d") and the angle of the ball's initial velocity (θ) express the initial height of the can.

(B) Solve for the travel time for the shot to cover that horizontal distance. (In terms of those variables.)

(C) Write an expression for the height of the ball at that time. (Measured from ground level.)

(D) Write an expression for the height of the can at that time. (Measured from ground level -- that's where the initial height of the can is needed.)

Compare!

 

[rudransh verma]

D) Write an expression for the height of the can at that time. (Measured from ground level -- that's where the initial height of the can is needed.)

I have written 3. I don’t understand how to write 4th
@berkeman help!

 

[Doc Al]

I have written 3. I don’t understand how to write 4th
@berkeman help!

So far, so good. For (C) you wrote the height as a function of time -- now plug in the time, which you found in step (B).

For (D), use the same approach but realize that you aren't starting from y = 0, but y = y₂ (which you found in step (A)).

 

[Doc Al]

For (D), use the most general form of the kinematic equation:
y = y₀ + v_0,y t + (1/2) a t²

You know y₀; you found that in (A). You know the initial velocity of the can. And you know the acceleration due to gravity.

 

[rudransh verma]

For (D), use the same approach but realize that you aren't starting from y = 0, but y = y2 (which you found in step (A)).

y3 in first box is vertical distance covered by ball whereas y3 in second box is the height of can after time t. For first y3 I took starting point as 0. For second y3 I took starting point as y2 and applied y3-y2= -1/2gt^2
But is it right to take final point as y3 and initial as y2 because y2>y3. Isn’t this negative displacement?

 

[Doc Al]

For second y3 I took starting point as y2 and applied y3-y2= -1/2gt^2

Good. Rearrange this to give the position of the can at time t: y3 = y2 -1/2gt^2

But is it right to take final point as y3 and initial as y2 because y2>y3. Isn’t this negative displacement?

The can is falling, after all. So its displacement from the starting point is negative. So what?

Now find its position at the time you calculated in step (B). And compare that position to that of the ball, which you should have done for step (C).

 

[rudransh verma]

Now find its position at the time you calculated in step (B). And compare that position to that of the ball, which you should have done for step (C).

Already done it. It’s shown in image above in the boxes. Both y3s at time t are same.
So at the same time t both the y3 are same. This means position of the can and ball are same after time t from below and above. If both y3 had different values they would not have hit each other.

 

[Doc Al]

Excellent.

 

[rudransh verma]

Excellent.

So y2 and origin are fixed but y3 is not. To make it more clear I could have taken y31 and y32.
Thank you
But I have a question. We took the initial velocity of can to be zero. Why have we taken initial velocity of ball not zero?

 

[Doc Al]

Why have we taken initial velocity of ball not zero?

The ball is being shot at the can. If its initial velocity were zero, it would just sit there. (It's already on the ground.) On the other hand, the can is being dropped from rest.

 

[rudransh verma]

The ball is being shot at the can. If its initial velocity were zero, it would just sit there. (It's already on the ground.) On the other hand, the can is being dropped from rest.

So when we do something against gravity we take initial velocity as nonzero of the body but when we leave body in gravity we say it started from rest?

 

[Doc Al]

So when we do something against gravity we take initial velocity as nonzero of the body but when we leave body in gravity we say it started from rest?

It all depends on the problem you're trying to analyze. In this problem, we are shooting a projectile at a target that is dropped from rest.