What Is the Minimum Delay Time for Two Projectiles to Collide Mid-Air?

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Suk-Sci
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Two balls problems...please help!

Homework Statement


Two particles are thrown vertically upwards along the same line with the same initial velocity u=30m/s, but at different times. The second particle is thrown t seconds after the first. What should be the minimum value of t so that the particles collide in mid-air? Neglect air resistance.

Homework Equations


2as=v^2-u^2
v=u+at
s=ut+1/2at^2

The Attempt at a Solution


the distance covered by ball1 before lauch of ball 2 = 30t-5t^2
ball1 will stop 45m above the ground in 3s
not getting the next part...
 
on Phys.org


When the two balls collide, they must have the same displacement.
So for first ball y = ut - (1/2)gt^2...(1)
For second ball y = u(t-1) - (1/2)g(t-1)^2...(2)
Equate the two equations and solve for t.
 


Suk-Sci said:
Why(t-1)?
Because the second ball travels less time in air before collision, and the time difference is 1 s as it is projected one second later.
 


The second ball is thrown up t time later than the first one. If you measure the "running time" T from the instant when the second ball is thrown, the positions of the balls as function of T are:

y1=30 (t+T)-0.5 g (T+t)2

y2=30 T-0.5 g T2.

They meet when y1=y2.

ehild
 


Gravity is same for both particles, so doesn't this mean any t would satisfy the answer as long as t < the time taken for first ball to go up and come back down?

Im reasoning this because they are going to take the exact same path and hence even if u throw it just a nanosecond later, theyre going to collide at the turning point when the first ball starts having negative velocity and the 2nd ball is still rising to occupy the peak y position that it is in.