Answer :

Consider Ball 1:

At t = 0, Ball 1 is thrown downward with an initial velocity of 0.

**Symbols Used:**

u = Initial Velocity; v = Final Velocity; s = Distance Travelled; a = Acceleration

For Ball 1:

u = 0

t = 18 – 0 = 18 s

s = ?

a = g (Acceleration due to gravity)

Now, we know from the equations of motion that:

s = 1587.6 m

Now, for Ball 2:

u = v

t = 18 – 6 = 12 s (Since Ball 2 is dropped at t = 6 s)

s = same as for Ball 1, since both meet at same point at t = 18 s

a = g

Now, using equations of motion:

v = 7.35 ms^{–1}

Hence, the speed of Ball 2 must be 7.35 ms^{–1} to meet Ball 1 after 18 s.

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