Question

A ball is dropped from a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t=18 s.What is the value of v ?

Answer 1

for first ball :
velocity of ball after 6 sec = - gt = -60 m/s displacement covered by first ball = -1/2gt²
= -1/2 × 10 × (6)² = -180 m

for 2nd ball :
inital velocity of ball = -v m/s

both meet at t = 18 sec
use relative kinematic equation ,
Srel = Sorel + Urel.t +1/2arelt²
here,
Srel = 0 { because finally they meet }
Sorel = first ball position with respect to second ball position = ( 0 - (-180)) = 180m
Urel = ( -60 - (-v)} = (60 - v)
hence,
0 = 180 + (60 - v)T
180 = (V-60)×18
10 + 60 = v
v = 70 m/s



method 2 :-
distance covered by first ball in 6 sec
= 180m
speed of first ball = gt = 60 m/s

T = distance between them /( relative velocity)
because both move same direction so,
Vrel = ( v - 60)

18 = 180/(v - 60)
v = 70 m/s