# Basic kinematics : 1-Dimensional collision in mid-air

• Matty R
In summary, the conversation discusses a problem where two stones are projected vertically upwards from the same point with the same speed, and the goal is to find where they will meet. The conversation includes equations and attempts at solving the problem, ultimately resulting in a solution of 0.55m for the meeting point of the two stones.
Matty R
Hello.

I've been stuck on this question for ages, and was hoping someone could help me with it.

## Homework Statement

A stone is projected vertically upwards with a speed of 6m/s, and one second later a second stone is projected vertically upwards from the same point with the same speed.

Find where the two stones meet.

## Homework Equations

http://img193.imageshack.us/img193/4696/4bes.jpg

## The Attempt at a Solution

http://img195.imageshack.us/img195/908/4bp1.jpg

http://img140.imageshack.us/img140/1691/4bp2.jpg

http://img193.imageshack.us/img193/4383/4bp3.jpg

http://img148.imageshack.us/img148/6464/4bp4.jpg

As far as I can tell, I've worked out how high above the projection point the first stone is, when the second stone is launched.

I also worked out that the first stone is traveling at 4m/s at 1m above the ground after 1 second.

I tried doing this through trial and error, and came up with the collision at 0.15m above the projection point, 0.17 seconds after the second stone is launched, but I don't think that is the method I'm supposed to be using here. :shy:

I don't think I've ever done a question like this before, so I would really appreciate any help.

Thanks.

Last edited by a moderator:
for first particle $$$s_1 = ut_1 - \frac{1}{2}gt_1^2$$$
for second particle $$$s_2 = ut_2 - \frac{1}{2}gt_2^2$$$
we know that $$$t_1-t_2=1 , t_2=t_1-1$$$
then $$$s_2 = u(t_1-1) - \frac{1}{2}g(t_1-1)^2$$$.
since they meet, s_1 should be equal to s_2.
$$$s_1 = ut_1 - \frac{1}{2}gt_1^2 = u(t_1-1) - \frac{1}{2}g(t_1-1)^2=s_2$$$
after solving the equation above for u=6m/s and g=10m/s^2, we find $$$t_1=1.1s ,t_2=0.1s$$$ and $$$s_1=s_2=0.55m$$$ .They meet at 0.55m above the ground.

boris1907 said:
for first particle $$$s_1 = ut_1 - \frac{1}{2}gt_1^2$$$
for second particle $$$s_2 = ut_2 - \frac{1}{2}gt_2^2$$$
we know that $$$t_1-t_2=1 , t_2=t_1-1$$$
then $$$s_2 = u(t_1-1) - \frac{1}{2}g(t_1-1)^2$$$.
since they meet, s_1 should be equal to s_2.
$$$s_1 = ut_1 - \frac{1}{2}gt_1^2 = u(t_1-1) - \frac{1}{2}g(t_1-1)^2=s_2$$$
after solving the equation above for u=6m/s and g=10m/s^2, we find $$$t_1=1.1s ,t_2=0.1s$$$ and $$$s_1=s_2=0.55m$$$ .They meet at 0.55m above the ground.

Thank you so much for this. It's exactly what I was after. I haven't done anything like this before, so it took some time to get my head around it, but I think I understand it all now. I managed to work through it all and end up with 0.55 for s1 and s2.

Thanks again.

## 1. What is basic kinematics?

Basic kinematics is the branch of physics that studies the motion of objects without considering the forces that cause the motion. It focuses on describing the position, velocity, and acceleration of an object over time.

## 2. What is a 1-dimensional collision?

A 1-dimensional collision is a type of collision where two objects, moving along the same straight line, collide with each other. This means that the motion of the objects before and after the collision can be described using only one dimension, typically the x-axis.

## 3. What is mid-air collision?

A mid-air collision is a type of collision that occurs between objects that are in motion through the air. This can happen between two objects in free fall, such as a skydiver and their parachute, or between two moving objects in the air, such as two airplanes.

## 4. How is 1-dimensional collision in mid-air different from other types of collisions?

A 1-dimensional collision in mid-air is different from other types of collisions because it involves objects that are moving through the air. This means that factors such as air resistance and the objects' trajectories must be taken into account when analyzing the collision.

## 5. How is the velocity of objects after a 1-dimensional collision in mid-air calculated?

The velocity of objects after a 1-dimensional collision in mid-air can be calculated using the conservation of momentum and the conservation of kinetic energy equations. These equations take into account the mass, velocity, and direction of the objects before and after the collision to determine the final velocities of the objects.

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