A ball thrown up vertically returns to the thrower after 6 s, find:
(a) The velocity with which it was thrown up,
(b) The maximum height it reaches, and
(c) Its position after 4 s.
Since the ball thrown up vertically returns to the thrower in 6 seconds, this means that the ball will take half of this time.
= 3 s
(a) Given,
Final velocity, v = 0
Initial velocity, u =?
Acceleration due to gravity, g = -9.8 m/s2 (As the ball goes up)
Time taken, t = 3 s
We know that,
v = u + gt
0 = u + (-9.8) × 3
0 = u – 29.4
u = 29.4 m/s
(b) Now, calculation of maximum height:
We know that,
v2 = u2 + 2gh
(0)2 = (29.4)2 + 2 × (-9.8) × h
0 = 864.36 – 19.6 h
19.6 h = 864.36
h = 
h = 44.1 m
(c) Finally, we have to calculate the position of ball after 4 seconds:
We know that,
h = ut +
gt2
h = 0×1+ 
×9.8 × (1)2 (Because t = 1 s)
= 0 + 4.9 ×1
= 4.9 m