A ball thrown up vertically returns to the thrower after 6s. Find (a) The velocity with which it was thrown up, (b) The maximum height it reaches, and (c) Its position after 4s.

The ball returns to the ground after 6 seconds. Thus the time taken by the ball to reach the maximum height (h) is 3 seconds i.e t=3s
Let the velocity with which it is thrown up be u
(a). For upward motion, v=u+at
∴0=u+(-10)×3
⇒u=30m/s
(b). The maximum height reached by the ball
h=ut+1/2at²
h=30×3+1/2(-10)×3²
h=45 m
(c). After 3 seconds, it starts to fall down.
Let the distance by which it falls in 1s be d
d=0+1/2at'² where t'=1s
d=1/2×10×(1)²=5 m
∴Its height above the ground, h'=45-5=40 m
Hence after 4 s, the ball is at a height of 40 m above the ground.