Answer:
t = o.6 s
Explanation:
Let ball thrown from below be A and ball dropped from above be B.
A and B meet when they both are same level above the ground. Then let A moved up a distance d and B dropped a distance h. Then you know
d + h = 15 m ---------------(1)
Now apply s = ut + [tex]\frac{1}{2}[/tex]at²
To A upwards,
d = 25t - [tex]\frac{1}{2}[/tex]gt² -----------------(2)
To B downwards,
h = 0 + [tex]\frac{1}{2}[/tex]gt² ----------------(3)
(1) = (2) + (3) ⇒ 15 = 25t
t = 0.6 s
