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

[Suk-Sci]

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...

 

[rl.bhat]

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]

Why(t-1)?

 

[rl.bhat]

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.

 

[ehild]

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

 

[Piyu]

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.