Solving for Time of Collision for 2 Particles with Initial Speed u

Click For Summary
Two particles are projected vertically upward from the same point with an initial speed u, and the time interval between their projections is T, where T is less than 2u/g. The challenge is to determine the time elapsed before they collide and their speeds just before impact. The distance formula x = -0.5gt^2 + ut is identified as unsuitable for finding the collision height, as it represents distance traveled rather than height. To find the collision time, the equations for displacement of both particles should be equated, taking into account the time for the first particle as t and for the second as t-T. By solving for t in terms of T, the problem can be addressed effectively.
Jess1986
Messages
43
Reaction score
0
This is a follow from a post i made yesterday. The question is:

2 particles are projected vertically up one after another from the same point at same initial speed u. T is the time interval between the 2 projections, where T<2u/g. Find what further time elapses before the particles collide, and find their speeds immediately before impact.

The equations i have found so far are for distance traveled
x= -0.5gt^2 + ut.
However i have realized this formula is for the distance the particle has traveled rather then the height reached. Since i need to find the time when the particles will collide (ie when they are at the same height) i am stuck as i think the formula is no use?! Please help
 
Physics news on Phys.org
Hint : x is not actual distance but displacement.
Can u go from here?
 
so i should equate the equations for x for both particles? how would i substitute T into the equation for particle 2?
 
Yes u should equate x.

Take t as time for first proj. and t-T for second proj.
Solve for t in terms of T.
 
ok that's great thanks for your helpx
 

Thread 'Magnitude of buoyant force in fluids of different densities'

The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?
View full post »

Similar threads

Calculate the speed and angle of the center of mass before a collision
  • · Replies 3 ·
Replies
3
Views
992
A ball is thrown w/initial speed vi at an angle 𝜃i with the horizontal
  • · Replies 5 ·
Replies
5
Views
1K
Help with question on motion: Avoiding a rear-end collision
  • · Replies 6 ·
Replies
6
Views
3K
Motion puzzle: Bird flying between a train and a platform
  • · Replies 84 ·
3
Replies
84
Views
5K
Minimum initial speed to spin a particle around a disk (with gravity)
  • · Replies 9 ·
Replies
9
Views
2K
Solved problem: Projectile in slanting tunnel
  • · Replies 6 ·
Replies
6
Views
1K
Solving the Elastric 3-Body Collision Problem
  • · Replies 2 ·
Replies
2
Views
3K
Motion of center of mass under gravity
  • · Replies 10 ·
Replies
10
Views
2K
What Was the Initial Speed of the Proton in a Collision?
  • · Replies 2 ·
Replies
2
Views
3K
Closest distance of approach between 2 charged particles
  • · Replies 5 ·
Replies
5
Views
1K