Where Do Falling and Projected Stones Meet?

[Yashbhatt]

Homework Statement

This is a very simple question. I even got the answer but I don't exactly I got it. Here it is:
A stone is allowed to fall from the top of a 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where will the two stones meet.

Homework Equations

v² - u² = 2as

s = ut + 1/2at²

v = u + at

 

[adjacent]

Homework Statement

This is a very simple question. I even got the answer but I don't exactly I got it. Here it is:
A stone is allowed to fall from the top of a 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where will the two stones meet.

Homework Equations

v² - u² = 2as

s = ut + 1/2at²

v = u + at

It will meet when the distance from the ground of two stones become equal.
Calculate that.

 

[Yashbhatt]

It will meet when the distance from the ground of two stones become equal.
Calculate that.

We can also say that that they will meet at a point where the sum of distances covered by both the stones will 100. Is that correct?

 

[adjacent]

We can also say that that they will meet at a point where the sum of distances covered by both the stones will 100. Is that correct?

No.Do it like this:

How is the distance traveled by the thrown stone determined?
That distance should be equal to the distance between the falling stone and the ground.So it's 100- the distance traveled by the falling stone.

So you should use the equation relating displacement,time,initial velocity and acceleration(g in this case).

 

[Yashbhatt]

Let A be the point from where the first stone is dropped. Let B be the point from where the second stone is projected. Let the balls meet at P after time t has elapsed. So, then AP + PB = 100. Let v and u be the speeds of the first and second stone respectively. So,
1/2gt² + ut - 1/2gt² = 100
So, ut = 100 and we get t = 4s.

Is this correct?

 

[nasu]

It's OK. You can do it this way too.

Actually this shows you that the two stones have a uniform relative motion.
As they both move with the same acceleration, their relative acceleration is zero.

 

[Yashbhatt]

@nasu Thanks. With this method I got the answer as 4s and distance 20m. Is the answer correct?

 

[adjacent]

@nasu Thanks. With this method I got the answer as 4s and distance 20m. Is the answer correct?

It's correct.
My method:
$100-\frac{1}{2}gt^2=25t+\frac{1}{2} \times -gt^2$ also gives t=4s s=20m.

EDIT: I just realized that this is the same thing :shy:

 

[Yashbhatt]

OK . Thanks.