Two particles are thrown vertically upward at interval of 2 second with an initial velocity of 4 m/s each. Find the time when they will meet each other and also

Question

Two particles are thrown vertically upward at interval of 2 second with an initial velocity of 4 m/s each. Find the time when they will meet each other and also find the distance from the ground where they will meet ?

Answer 1

Tony,

 

There is a problem with this question as it is asked, though it is easily solvable.  Using the equations of motion we can get a solution:

 

y = y ₀ + v _0*t + ½ a _y*t ² 

 

and 

 

v = v ₀ + a*t

 

We would simply need to write this equation twice, once for each particle, with the condition for particle 2 that t ₂ = t ₁ + 2 sec.

 

Here's the "catch":  the initial velocity of each particle is only 4 m/s!  Under the influence of gravity =

9.8 m/s ² , a particle will gain or lose this much velocity in a time t = (v-v ₀)/a .

 

The total time for the particle to reach the apex (top) of its trajectory and return to the ground will be 2*t.

 

Let's solve for t at the top of the trajectory.  The velocity at the top of the trajectory for the particles will be zero, so:

 

with a = g = -9.8 m/s ² and using v = v ₀ + a* t gives:

 

t = ( 0 - 4 m/s)/(-9.8 m/s ²⁾ = 0.408 seconds.  

 

Round trip time for the 1st particle is then 2*t = 0.816 sec.

 

Since the second particle is only being thrown at t + 2 sec, the first particle lands about 1.2 sec before the second particle is thrown.

 

Your answer to the question as stated is then t = 0.816 sec and d = 0!