Question -
Answer -
(a)Time of ascent is equal to the time of descent. The ball takes a total of 6 sfor its upward and downward journey.
Hence,it has taken 3 s to attain the maximum height.
Final velocity of the ball at the maximumheight, v = 0
Acceleration due to gravity, g = −9.8 m s−2
Equation of motion, v = u + gt willgive,
0 = u + (−9.8 × 3)
u = 9.8 × 3 = 29.4 ms−1
Hence, the ball was thrown upwards with avelocity of 29.4 m s−1.
(b) Let the maximum height attained by theball be h.
Initial velocity during the upwardjourney, u = 29.4 m s−1
Final velocity, v = 0
Acceleration due to gravity, g = −9.8 m s−2
Fromthe equation of motion,
(c)Ball attains the maximum height after 3 s. After attaining this height, it willstart falling downwards.
Inthis case,
Initial velocity, u = 0
Positionof the ball after 4 s of the throw is given by the distance travelled by itduring its downward journey in 4 s − 3 s = 1 s.
Equationof motion, 
willgive,Totalheight = 44.1 m
Thismeans that the ball is 39.2 m (44.1 m − 4.9 m) above the ground after 4seconds.