Doubt by Krish
Solution :
Let both the balls will collide after time t at point C which is at a height x from the bottom.
Distance between both the balls = AB = 40 m
AC = x m
BC = (40-x) m
Initial velocity with which both the balls are thrown upward = 20 m/s
For Ball A
S=ut +½at²
x=20t+½(-g)t²
x=20t-½gt² — (1)
x=20t-½gt² — (1)
For Ball B
S=ut +½at²
40-x=20t+½(g)t²
40-x=20t½gt² — (2)
40-x=20t½gt² — (2)
Solving equation (1) and (2)
x=20t-½gt²
40-x=20t½gt²
----------------------
40=40t + 0
----------------------
t=40/40
t=1 sec
Putting t=1sec in equation (1)
x=20(1)-½(10)(1)²
x=20-5
x=15 m
x=15 m
AC= 15 m
BC = 40-x
BC = 40-15
BC = 25 m
BC = 40-15
BC = 25 m
Hence, both the balls will collide each other after 1 sec at a distance of 15 m from the bottom or 25 m from the top.