A ball is thrown vertically upwards. It returns 6 s later. Calculate (i) the greatest height reached by the ball, and (ii) the initial velocity of the ball. (Take g = 10 m s-2 )

(i) Initial velocity (u) = 0

g = 10 m s^-2

Since, the total time for the journey of the ball in the air = 6 s.

Therefore, time for reaching maximum height = 6/2 = 3 s

h = ut + 1/2 gt²

= 0 × 3 + 1/2 × 10 × 3²

= 0 + 5 × 9 = 45 m

Hence, greatest height reached by the ball = 45 m

(ii) From the equation of motion,

v² = u² – 2gs

⇒ 0² = u² – (2 × 10 × 45)

⇒ u² = 900

⇒ u = √900 = 30 m s^-1

Hence, the initial velocity of the ball = 30 m s^-1